CHM130 Summary Final Editing PDF

DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 Chapter – 1:  Significant figure  Conversion Chapter-2  Atomic & molecular mass  Mole concept and avogadro’s number  Calculation involving formula and atomic mass  Percentage by mass of an element in a compound  Empirical...

DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 Chapter – 1:  Significant figure  Conversion Chapter-2  Atomic & molecular mass  Mole concept and avogadro’s number  Calculation involving formula and atomic mass  Percentage by mass of an element in a compound  Empirical and molecular formula  Balancing of chemical equations  Calculation involving reacting mass  Calculation involving solution preparation  Calculation involving reacting solutions  Calculation of yields  Chemical reaction  Types of chemical reaction Chapter-3:  Basic ideals about the atomic structure  Isotopes  Quantum number  Electron configuration of elements Chapter-4:  Periodic table Chapter-5:  Chemical bonding  Bond polarity  Octet rule  Lewis structure  -failure of the octet rule  Formal charge  Resonance  Miscellaneous Chapter -6  Molecular orbital theory (MOT) Chapter- 7:  Thermodynamics  Thermochemistry  Standard states and enthalpy of reaction  Hess’s law  Bond energy  Entropy Chapter – 8: Gases Like our facebook page: facebook.com/DelzyScholars 1|P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 SIGNIFICNT FIGURE (S.F) The term significant figures refer to the number of important single digit (0 through 9 inclusive) in a number expression. RULES FOR SIGNIFICNT FIGURES NOTE: All the numbers used as examples are assumed to be a measured quantity. 1. All non-zero digits (1 through 9) ARE significant: Example; 33.2m has 3 S.F; 44.56m has 4 S.F; 55.367m has 5 S.F Note: Zero digits may count or may not count as been significant depending on their position in a number expression, and as such we will be looking at the various types of zeros that can exist in a number. 2. CAPTIVE ZEROS: These are zeros between two non-zero digit. CAPTIVE zeros ARE significant. (202), (80.085), etc, the underlined zeros are captive zeros. Example:  2051 contains one (1) captive zero (underlined) and 3 non-zero digit, therefore 2051 contains 4 S.F  200.35 contains two (2) captive zeros (underlined) and 3 non-zero digits, therefore 200.35 contains 5 S.F. N/b: Captive zeros are significant 3. LEADING ZEROS: These are zeros that precedes all non-zero digits. They ARE NOT significant. (0.23), (0.0045), etc, the underlined zeros are leading zeros. Example:  0.54 contains one (1) leading zero (underlined) and 2 non-zero digit, therefore 0.54 contains 2 S.F, since leading zeros are not always significant.  0.0032 contains one (3) leading zero (underlined) and 2 non-zero digit, therefore 0.0032 contains 2 S.F, since leading zeros are not always significant.  0.323 contains one (1) leading zero (underlined) and 3 non-zero digit, therefore 0.323 contains 3 S.F, since leading zeros are not always significant. N/b: Leading zeros are not significant 4. TRAILING ZEROS: These are zeros at the right end of numbers. They ARE significant ONLY if the number contains a decimal point. E.g 25.00 (contains two (2) trailing zeros and they are significant since the number contains a decimal point. Further Example:  92.00 contains two (2) trailing zero (underlined), and the number expression contains a decimal point and as such the zeros ARE significant, plus the two(2) non-zero digit (92), therefore 92.00 contains 4 S.F  9200 contains two (2) trailing zeros (underlined), and the number expression do not contain a decimal point and as such the zeros ARE NOT significant, the two(2) non-zero digit (92) are significant, therefore 9200 contains 2 S.F Like our facebook page: facebook.com/DelzyScholars 2|P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576  5000 contains three (3) trailing zeros (underlined), and the number expression does not contain a decimal point and as such the three (3) trailing zeros ARE NOT significant, leaving us with digit (5) as the only significant figure. There 5000 contains 1 S.F.  5.000 contain three (3) trailing zeros and also do contain a decimal point and as such the three (3) trailing zeros ARE significant plus one (1) non zero digit (5). Therefore 5.000 contains 4 S.F For a number in scientific notation; N × 10˟, all digits comprising N are significant by the first four rules above. Example; 5.02 × 104;  N = 5.02; and as such contains 3 S.F 1.102 × 104;  N = 1.102; and as such contains 4 S.F Note: 5000 if expressed in scientific notation can have 1, 2, 3 and 4 significant figure provided it is expressed in standard form 5 × 103;  N = 5; and as such contains 1 S.F. Therefore; (5 × 103 has 1 S.F) 5.0 × 103;  N = 5.0; and as such contains 2 S.F. Therefore; (5.0 × 103 has 2 S.F) 5.00 × 103;  N = 5.00; and as such contains 3 S.F. Therefore; (5.00 × 103 has 3 S.F) 5.000 × 103;  N = 5.000; and as such contains 4 S.F Therefore; (5.000 × 103 has 4 S.F) Problem examples (PAST QUESTIONS) 1. How many significant figures are in the quantity 1.0094? Solution 1.0094? Contains the following; Non-zero digits; ARE significant =3 Captive zeros; ARE significant = 2 Therefore 1.0094 contains 5 S.F 2. How many significant figures are in the quantity 0.00330? (assume that it’s a measured quantity) Solution 0.00330 Contains the following; Non-zeros digit; ARE significant =2 Leading zeros; ARE NOT significant = 3 Trailing zeros: ARE significant if decimal point is present = 1 Therefore 0.00330 contains 3 S.F (2+1) 3. Underline the zeros that are significant figures and determine the number of significant figures in the quantity 0.1044. Solution 0.1044 Contains the following; Captive zeros (underlined): ARE significant =1. Leading zeros ARE not significant =1 Non-zero digits ARE significant =3 Therefore; 0.1044 Contains 1+3 =4 S.F Like our facebook page: facebook.com/DelzyScholars 3|P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 CONVERSION: Conversion is a process of changing from one unit to another using appropriate conversion factor. Useful conversion factors and relationships LENGTH: S.I unit for length is meter (m) 1km 0.6217miles (mi) 1mile (mi) 5280ft 1mile (mi) 1.6093km 1 meter (m) 1000 millimeters (mm) 1 meter (m) 100 centimetres (cm) 1 meter (m) 10 decimetres (dm) 1 meter (m) 3.281 feet 1 meter (m) 1.094 yards 1 yard 36 inches (in) 1 inch (in) 2.54cm 𝑜 1 Armstrong (𝐴 ) 10−10 𝑚 MASS: S.I unit for mass is kilogram (kg) 1kilogram (kg) 2.205 Ib 1 kilogram (kg) 1000 grams (g) 1 pounds (Ib) 453.59 grams (g) 1 gram 6.022 × 1023 𝑎𝑚𝑢 VOLUME: S.I unit for volume is cubic meter (𝑚3 ) 1 Litres (L) 1 cubic decimeters (𝑑𝑚3) 1 L or 𝟏 dm3 1000 cubic centimeters (𝑐𝑚3) 1 gallon (gal) 3.7854 Litres (L) Conversion factor expresses the relationship between two quantity having different units of measurement and can be written as a fraction in two forms. For example: 2.54 cm and 1 inch (in) are the same length, so we can say 2.54cm = 1 in. The relationship/equivalence between centimeters (cm) and inches (in) is 𝟐. 𝟓𝟒𝐜𝐦 = 𝟏 𝐢𝐧𝐜𝐡. We can write two conversion factors from this information. 2.54cm 1 inch or 1 inch 2.54cm The first will be use convert inches (in) to centimeters (cm) while the other is use to convert centimeters (cm) to inches (in) Example: How many centimeters are in 6.00 inches? 2.54cm 𝑥 cm = 6.00 in. × = 15.2cm 1 in. Therefore the length 6.00 inches is the same as the length 15.2cm Like our facebook page: facebook.com/DelzyScholars 4|P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 The unit inches in the denominator of the conversion factor cancels the unit inches in the given data (6.00 inches). The unit centimeters (cm) in the numerator of the conversion factor becomes the unit of the final answer Or you can still solve like follows: If 2.54 cm = 1 in. Therefore; 𝑥 cm = 6.00 in. Cross multiply and solve for x 2.54 cm × 6.00 in. 𝑥 cm = = 15.2cm 1 in. Generally we begin any conversion by examining the units of the given data and the units we desire. We then ask ourselves what is the equivalence/relationship between the unit of the given data and the unit we are asked to convert to. From the equivalence/relationship you will now know the conversion factor that would take us from the units of the given quantity to those of the desired one. USING TWO OR MORE CONVERSION FACTOR:  There are some cases whereby the direct relationship (conversion factor) between the given unit and desired won’t be given but rather two or more conversion factors that would lead us to getting the desired unit given. In this case you will carry out the conversion stepwise to arrive at the desired unit. Example: Convert the length of an 8.00m rod to inches. (2.54cm = 1 inch; 100cm = 1m) Solution: From the relationship given, we won’t be able to convert directly from meters (m) to inches (in). Therefore we would convert step by step: First by converting from meters to centimeters, and then from centimeters to inches as diagrammed 100 𝑐𝑚 1 𝑖𝑛. 𝑥 inches = 8.00 𝑚 × × = 𝟑𝟏𝟓 𝒊𝒏. 1𝑚 2.54 𝑐𝑚 The first conversion factor used is applied to cancel meters and convert the length to centimeters thus, meters are written in the denominator and centimeters in the numerator. The second conversion factor is written to cancel centimeters and convert the length to inches (desired unit). Or we can also solve to show each step as follows: 100 𝑐𝑚  First: convert from m to cm:  8.00 𝑚 × 1 𝑚 = 800𝑐𝑚 1 𝑖𝑛.  Second: convert the cm to inches:  800 𝑐𝑚 × = 𝟑𝟏𝟓 𝒊𝒏 2.54 𝑐𝑚 Therefore 8.00 meters (m) = 315 inches(in) Note: If this kind of question appears in exams, they may likely not give the conversion factor relating meters and centimeters reason is that there are some basic conversion factors that you should know and meters to centimeters is one of them.  In some other cases where two or more unit is to be converted to a desired new unit, two or more conversion factors will also be used. Like our facebook page: facebook.com/DelzyScholars 5|P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 Example: The density of benzene at 25o C is 0.879 𝑔/𝑐𝑚3. Convert this density to pounds per cubic inch (𝐼𝑏/𝑖𝑛3 ) using the following relationship (1 𝐼𝑏 = 453.59𝑔; 1 𝑖𝑛 = 2.54 𝑐𝑚) Solution: If you notice the conversion factor cm & inch given were in-terms of length so to get their cubic equivalent (volume), simply cube both sides. (1 𝑖𝑛)3 = (2.54 𝑐𝑚)3 (1)3 𝑖𝑛3 = (2.54)3 𝑐𝑚3 Note: (1)3 𝑖𝑛3 = 1 𝑖𝑛3 To go from the given units, 𝑔/𝑐𝑚3 , to the desired units 𝐼𝑏/𝑖𝑛3 , we will convert grams (g) to pounds (Ib) and then cubic centimetre (𝑐𝑚3 ) to cubic inch (𝑖𝑛3 ). Converting 𝟎. 𝟖𝟕𝟗 𝐠. 𝐜𝐦−𝟑 ===> 𝒙 𝐈𝐛/𝒊𝒏𝟑 we would have; 0.879 g 1 𝐼𝑏 2.54 𝑐𝑚 3 𝒙 = × × ( ) 𝑐𝑚3 453.59 𝑔 1 𝑖𝑛. 0.879 g 1 𝑰𝒃 16.387 𝑐𝑚3 𝒙 = × × = 𝟎. 𝟎𝟑𝟐 𝑰𝒃/𝒊𝒏𝟑 𝑐𝑚3 453.59 𝑔 1 𝒊𝒏𝟑 The first conversion factor used is applied to cancel grams and convert the mass to pounds (Ib) thus, grams (g) are written in the denominator and pounds (Ib) in the numerator. The second conversion factor is written to cancel cubic centimeters and convert the volume to cubic inch. Showing the solving step wise for further grasping: 1 𝐼𝑏  First: Convert 0.879 𝑔 𝑡𝑜 𝑝𝑜𝑢𝑛𝑑𝑠 (𝐼𝑏) : => 0.879 g × = 𝟎. 𝟎𝟎𝟏𝟗𝟒 𝑰𝒃 453.59 𝑔 1 𝑖𝑛3  Second: Convert 1.00 𝑐𝑚3 𝑡𝑜 𝑖𝑛3 : => 1.00𝑐𝑚3 × = 𝟎. 𝟎𝟔𝟏𝟎𝟐 𝒊𝒏𝟑 16.387 𝑐𝑚3  Third: Divide the mass in pounds (Ib) by the volume in cubic inch (𝑖𝑛3 ) 0.00194 𝑰𝒃 = 𝟎. 𝟎𝟑𝟐 𝑰𝒃/𝒊𝒏𝟑 0.06102 𝒊𝒏𝟑 Example - (Textbook exercise): If we assume that the density of a solution is 1.00 𝑔. 𝑐𝑚−3. Determine the volume of 150 pounds of the solution in litre. {1 pounds ( Ib) = 435.59 grams (g)} Solution: This is converting from g. cm−3 to Ib. L−1 (pounds per litre) We also know from our basic conversion that: 1 Ib = 435.59 g ; 1000 cm3 = 1 dm3 (L) Converting 𝟏. 𝟎𝟎 𝐠. 𝐜𝐦−𝟑 ===> 𝒙 Ib. L−1 we would have; 1.00 g 1 𝐼𝑏 1000 𝑐𝑚3 𝒙 = × × = 𝟐. 𝟐𝟎 𝐈𝐛. 𝐋−𝟏 𝑐𝑚3 453.59 𝑔 1𝐿 Therefore 1.00 g. cm−3 = 2.20 Ib. L−1 The first conversion factor used is applied to cancel grams and convert the mass to pounds (Ib) thus, grams (g) are written in the denominator and pounds (Ib) in the numerator. The second conversion factor is written to cancel cubic centimeters and convert the volume to litres (L). Like our facebook page: facebook.com/DelzyScholars 6|P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 2. Past question. Given the following information below: 2.54 cm = 1 in 12 in = 1 ft What is the price of a piece of a cloth 325cm long that sells for 15naira per foot (ft)? Solution: 15 naira 𝟏𝟓 𝒏𝒂𝒊𝒓𝒂 𝒑𝒆𝒓 𝒇𝒐𝒐𝒕 in terms of a conversion factor = ft 15naira First convert the 325cm to inches, then the inches to foot (ft) and then multiply with since per ft ft cost 15naira 1 in 1 ft 15 naira 325cm × × × = 𝟏𝟓𝟗. 𝟗𝟒 𝐧𝐚𝐢𝐫𝐚 2.54 cm 12 in ft Therefore 325cm length of the piece of the cloth will sell for 159.94 naira. Showing the breakdown solving (step wise): 1 in  First: Convert 325 cm to inches => 325cm × = 𝟏𝟐𝟕. 𝟗𝟓 𝐢𝐧 2.54 cm 1 ft  Second: Convert the inches to ft: => 127.95 in × = 𝟏𝟎. 𝟔𝟔𝟐𝟓 𝐟𝐭 12 in  Third: If 1 ft = 15 naira Therefore; 10.6625 ft = 𝒙 𝐧𝐚𝐢𝐫𝐚. 10.6625 ft × 15 naira 𝒙 𝐧𝐚𝐢𝐫𝐚 = = 𝟏𝟓𝟗. 𝟗𝟒 𝐧𝐚𝐢𝐫𝐚 1 ft OR for the third step: 15 naira We can simply multiply 10.6625 ft with to get the price of 10.6625ft piece of cloth ft since per foot is 15naira 15 naira 10. 6625 ft × = 𝟏𝟓𝟗. 𝟗𝟒 𝐧𝐚𝐢𝐫𝐚 ft 2. Past question. Convert the density of iron 7.9𝑔/𝑐𝑚3 𝑡𝑜 𝐼𝑏/𝑓𝑡 3 , using the relationship (1 𝐼𝑏 = 453. 59𝑔 ; 1 𝑖𝑛 = 2.54 𝑐𝑚). Leave the answer in two significant figures. Solution: If you notice the conversion factor cm & inch given were in-terms of length: 𝟏 𝒊𝒏 = 𝟐. 𝟓𝟒 𝒄𝒎 So to get their cubic equivalent (volume), simply cube both sides. (𝟏 𝒊𝒏)𝟑 = (𝟐. 𝟓𝟒 𝒄𝒎)𝟑 The conversion factor needed for this conversion isn’t complete, it is one short as we would need a relationship that will relate either of inches or centimeters to foot (ft) to enable us be able to the foot (ft). So we will get it from the conversion table. But from conversion table we know that 𝟏𝟐 𝐢𝐧. = 𝟏 𝐟𝐭 Like our facebook page: facebook.com/DelzyScholars 7|P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 So to get their cubic equivalent (volume), simply cube both sides. (𝟏𝟐 𝐢𝐧)𝟑 = (𝟏 𝐟𝐭)𝟑 To go from the given units, 𝑔/𝑐𝑚3 , to the desired units 𝐼𝑏/𝑓𝑡 3 , we will convert grams (g) to pounds (Ib) and then cubic centimetre (𝑐𝑚3 ) to cubic foot (𝑓𝑡 3 ). Converting the 𝟕. 𝟗 𝐠/𝐜𝐦𝟑 ===> 𝒙 𝐈𝐛/𝒇𝒕𝟑 we would have; 𝟕. 𝟗 𝒈 1 𝐼𝑏 2.54 𝑐𝑚 3 12 𝑖𝑛 3 𝒙 = × × ( ) × ( ) 𝒄𝒎𝟑 453.59 𝑔 1 𝑖𝑛. 1 𝑓𝑡 𝟕. 𝟗 𝒈 1 𝐼𝑏 16.387 𝒄𝒎𝟑 1728 𝒊𝒏𝟑 𝒙 = × × × = 𝟒𝟗𝟑. 𝟏𝟖 𝑰𝒃/𝒇𝒕𝟑 𝒄𝒎𝟑 453.59 𝒈 1 𝒊𝒏𝟑 1 𝑓𝑡 3 Since they said you express the answer in two (2) significant figures  𝟒𝟗𝟎 𝑰𝒃/𝒇𝒕𝟑 (𝟐 𝒔. 𝒇) The first conversion factor used is applied to cancel grams and convert the mass to pounds (Ib) thus, grams (g) written in the denominator and pounds (Ib) in the numerator. The second conversion factor is written to cancel the cubic centimeters (𝑐𝑚3 ) and convert the volume to cubic inch (𝑖𝑛3 ) while the third conversion factor is to cancel the cubic inch (𝑖𝑛3 ) and convert the volume to cubic foot (𝑓𝑡 3 ) which is the desired. OR Alternatively you can first convert the mass from grams (g) to pounds (Ib), then convert also the cubic centimeter (𝐜𝐦𝟑 ) to cubic foot (𝐟𝐭 𝟑 ) after which you divide the mass in pounds (Ib) by the volume in cubic foot (ft 3 ) 1 Ib − Converting 𝟕. 𝟗 𝐠 to pounds (Ib) : ==> 𝟕. 𝟗𝐠 × = 𝟎. 𝟎𝟏𝟕𝟒 𝐈𝐛 453.59 𝐠 1 𝑖𝑛. 3 1 𝑓𝑡 3 − Converting 1 𝐜𝐦𝟑 to cubic foot (𝐟𝐭 𝟑 ) : => 1 𝐜𝐦𝟑 × ( ) × ( ) = 𝟑. 𝟓𝟑 × 𝟏𝟎−𝟓 𝐟𝐭 𝟑 2.54 𝑐𝑚 12 𝑖𝑛 Divide the mass (𝟎. 𝟎𝟏𝟕𝟒 𝐈𝐛) by the volume (𝟑. 𝟓𝟑 × 𝟏𝟎−𝟓 𝐟𝐭 𝟑 ): 0.0174 𝐈𝐛 = 493 𝐈𝐛/𝐟𝐭 𝟑 3.53 × 10−5 𝐟𝐭 𝟑 In two (2) significant figures  𝟒𝟗𝟎 𝑰𝒃/𝒇𝒕𝟑 (𝟐 𝒔. 𝒇) NOTE: TEMPERATURE CONVERSION DOES NOT FOLLOW THIS SAME PATTERN. Like our facebook page: facebook.com/DelzyScholars 8|P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 FORMULA WEIGHT & MOLECULAR/MOLAR MASS The formula weight/mass of a substance is the sum of the atomic weights/masses of each atom in its chemical formula. If the chemical formula is that of molecule (molecular formula), then the formula mass is called molar mass or molecular mass The molar mass/molecular mass of any substance is always numerically equals to its formula mass. Molar mass/molecular mass is the mass in grams of one mole of a substance (molecule or compound). Hence one can calculate the formula mass of a substance from the formula of the substance by adding up the atomic masses of each element that makes up the formula. Similarly, the molecular/molar mass of a substance can be calculated from the molecular formula of one mole of the substance by adding up the atomic masses of each of the component atoms making up the molecular formula Atomic mass is simply the mass in gram of a single atom. Note: the unit of molar mass = g/mol Problem-1: Determine the molar mass of hydrated copper sulphate salt with molecular formula 𝐶𝑢𝑆𝑂4. 5𝐻2 𝑂. (Relative atomic masses of atoms are Cu = 63.5, S = 32, O = 16, H =1.0) Solution: Molar mass of 𝐶𝑢𝑆𝑂4. 5𝐻2 𝑂 = 𝐶𝑢 + 𝑆 + 4𝑂 + 5{2𝐻 + O} {(𝟔𝟑. 𝟓) + (𝟑𝟐) + (𝟏𝟔 × 𝟒) + 𝟓{(𝟐 × 𝟏) + 𝟏𝟔} = 𝟐𝟒𝟗. 𝟓𝒈/𝒎𝒐𝒍 This implies that the mass in 1mole of 𝐶𝑢𝑆𝑂4. 5𝐻2 𝑂 = 𝟐𝟒𝟗. 𝟓𝒈 Problem-2: Determine the molar mass of hydrogen tetraoxosulphate VI acid with molecular formula 𝑯𝟐 𝑺𝑶𝟒. (Relative atomic masses of atoms are, S = 32, O = 16, H =1.0) Solution: Molar mass of 𝑯𝟐 𝑺𝑶𝟒 = {2H + S + 4O} {(𝟐 × 𝟏. 𝟎) + (𝟑𝟐) + (𝟒 × 𝟏𝟔)} = 𝟗𝟖𝒈/𝒎𝒐𝒍 This implies that the mass in 1mole of 𝑯𝟐 𝑺𝑶𝟒 = 𝟗𝟖𝒈 Problem-3: (Complete the solving..) Determine the molar mass of a molecule with molecular formula 𝐂𝟔 𝐇𝟏𝟐 𝐎𝟔 (Relative atomic masses of atoms are, C = 12, O = 16, H =1.0) Like our facebook page: facebook.com/DelzyScholars 9|P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 PERCENTAGE (%) COMPOSITION OF A COMPOUND The percent composition of a compound is the mass percent contributed by each element in the compound. The percentage by mass of each element present in a compound can be worked out using the formula: % 𝒃𝒚 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒂𝒏 𝒆𝒍𝒆𝒎𝒆𝒏𝒕 𝒊𝒏 𝒂 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅 𝑛 × 𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 = × 100% 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 Problem example: Find the percentage by mass of each element present in 𝑪𝟔 𝑯𝟓 𝑵𝑶𝟐 (C=12, H=1, N=14, O=16) Solution: The relative molecular mass/molar mass of 𝑪𝟔 𝑯𝟓 𝑵𝑶𝟐 = (𝟏𝟐 × 𝟔) + (𝟏 × 𝟓) + (𝟏𝟒) + (𝟏𝟔 × 𝟐) = 𝟏𝟐𝟑𝒈/𝒎𝒐𝒍 𝟔 ×𝟏𝟐  (𝐢) % 𝑏𝑦 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑪 𝑖𝑛 𝑡ℎ𝑒 𝑪𝟔 𝑯𝟓 𝑵𝑶𝟐 = 𝟏𝟐𝟑 × 𝟏𝟎𝟎 = 𝟓𝟖. 𝟓𝟒% 𝟓 ×𝟏  (ii) % 𝑏𝑦 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑯 𝑖𝑛 𝑡ℎ𝑒 𝑪𝟔 𝑯𝟓 𝑵𝑶𝟐 = × 𝟏𝟎𝟎 = 𝟒. 𝟎𝟔% 𝟏𝟐𝟑 𝟏 ×𝟏𝟒  (𝐢𝐢𝐢)% 𝑏𝑦 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑵 𝑖𝑛 𝑡ℎ𝑒 𝑪𝟔 𝑯𝟓 𝑵𝑶𝟐 = × 𝟏𝟎𝟎 = 𝟏𝟏. 𝟑𝟖% 𝟏𝟐𝟑 𝟐 ×𝟏𝟔  (𝐢𝐯) % 𝑏𝑦 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑶 𝑖𝑛 𝑡ℎ𝑒 𝑪𝟔 𝑯𝟓 𝑵𝑶𝟐 = × 𝟏𝟎𝟎 = 𝟐𝟔. 𝟎𝟐% 𝟏𝟐𝟑 Note: addition of the percentage composition of the elements that makes up a compound would sum up to 100% Problem example: A compound with the molecular formula 𝐶𝑢𝑆𝑂4. 5𝐻2 𝑂 (hydrated copper sulphate salt)  (a) Determine the percentage by mass of all the elements in the compound.  (b) Calculate the percentage of water in the compound Solution: (Cu = 63.5, S = 32, O = 16, H =1) Relative molecular mass/molar mass of 𝐶𝑢𝑆𝑂4. 5𝐻2 𝑂 = {(𝟔𝟑. 𝟓) + (𝟑𝟐) + (𝟏𝟔 × 𝟒) + (𝟓 × 𝟏𝟖)} = 𝟐𝟒𝟗. 𝟓𝒈/𝒎𝒐𝒍 (a) Determine the percentage by mass of all the elements in the compound. 𝟏 × 𝟔𝟑.𝟓  (𝒊) % 𝑪𝒖 𝒊𝒏 𝒕𝒉𝒆 𝐶𝑢𝑆𝑂4. 5𝐻2 𝑂 = 𝟐𝟒𝟗.𝟓 × 𝟏𝟎𝟎 = 𝟐𝟓. 𝟒𝟓% 𝟏 × 𝟑𝟐  (𝒊𝒊) % 𝑺 𝒊𝒏 𝒕𝒉𝒆 𝐶𝑢𝑆𝑂4. 5𝐻2 𝑂 = × 𝟏𝟎𝟎 = 𝟏𝟐. 𝟖𝟑% 𝟐𝟒𝟗.𝟓 𝟗 × 𝟏𝟔  (𝒊𝒊𝒊) % 𝑶 𝒊𝒏 𝒕𝒉𝒆 𝐶𝑢𝑆𝑂4. 5𝐻2 𝑂 = × 𝟏𝟎𝟎 = 𝟓𝟕. 𝟕𝟐% 𝟐𝟒𝟗.𝟓 𝟏𝟎 × 𝟏  (𝒊𝒗) % 𝑯 𝒊𝒏 𝒕𝒉𝒆 𝐶𝑢𝑆𝑂4. 5𝐻2 𝑂 = × 𝟏𝟎𝟎 = 𝟒. 𝟎𝟎% 𝟐𝟒𝟗.𝟓 (b) Calculate the percentage of water in the compound In this case you are asked to find the percent that water contributed to the total mass of the compound.  𝐂𝐮𝐒𝐎𝟒. 𝟓𝐇𝟐 𝐎 contains 5 molecules of water; 𝒏 = 𝟓; 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐻2 𝑂 = 𝟏𝟖𝐠/𝐦𝐨𝐥 𝟓 × 𝟏𝟖 % 𝑯𝟐 𝑶 𝑖𝑛 𝑡ℎ𝑒 𝑪𝒖𝑺𝑶𝟒. 𝟓𝑯𝟐 𝑶 = × 𝟏𝟎𝟎 = 𝟑𝟔. 𝟎𝟕% 𝟐𝟒𝟗. 𝟓 Like our facebook page: facebook.com/DelzyScholars 10 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 EMPIRICAL AND MOLECULAR FORMULA: Empirical formula: Empirical formula of a compound is the simplest formula of the compound showing the relative mole ratio of each type of atoms present in the compound. It only shows the proportion in which the component elements are present in a compound. Note: Chemical analysis of a substance can provide the composition of the substance by mass or by percentage. Problem example: What is the empirical formula of L-Dopa, a drug used for the treatment of Parkinson’s disease if it has the following composition: 54.82% C, H, 7.10% N, and 32.46% O? Solution: C = 54.82%, N = 7.10%, O = 32.46% H = 100 % − (54.82 + 7.10 + 32.46)% = 5.62% C H N O % Mass 54.82 5.62 7.10 32.46 Step-1: 𝟓𝟒.𝟖𝟐 𝟓.𝟔𝟐 𝟕.𝟏𝟎 𝟑𝟐.𝟒𝟔 Divide by the atomic 𝟏𝟐 𝟏 𝟏𝟒 𝟏𝟔 mass of the elements = 𝟒. 𝟓𝟕 = 5.62 = 0.51 = 2.03 Step-2: 𝟓.𝟔𝟐 𝟎.𝟓𝟏 𝟐. 𝟎𝟑 𝟒.𝟓𝟕 Divide through the 0.51 0.51 0.51 0.51 answers obtained by the smallest = 9.0 = 11 =1 =4 value Therefore the Empirical formula = 𝑪𝟗 𝑯𝟏𝟏 𝑵𝑶𝟒 2. What is the empirical formula of a compound that contains 0.783g of carbon, 0.196g of Hydrogen and 0.521g of oxygen? (C=12, O=16, H=1) C H O Mass (g) 0.783 0.196 0.521 Step-1: 0.783 0.196 0.521 Divide by the atomic 12 1 16 mass of the elements 𝟎. 𝟎𝟔𝟓 0.196 0.033 Step-2: 𝟎.𝟎𝟔𝟓 𝟎.𝟏𝟗𝟔 𝟎. 𝟎𝟑𝟑 Divide through the answers obtained 0.033 0.033 0.033 by the smallest = 2 = 6 =1 value Therefore the Empirical formula = 𝐂𝟐 𝐇𝟔 𝐎 Like our facebook page: facebook.com/DelzyScholars 11 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 2. What is the empirical formula of a compound that contains 89.14% Au and 10.80% of O? (Au=197, O=16) Au O % Mass 89.14 10.86 Step-1: 89.14 10.86 Divide by the atomic 197 16 mass of the elements 0.45 0.68 Step-2: 𝟎.𝟒𝟓 𝟎. 𝟔𝟖 Divide through the answers obtained 0.45 0.45 by the smallest = 1 = 1.5 value Once you have.5 (point 5) in your final answer, multiply the whole values by 2 Multiplying by 2 2 3 Therefore the Empirical formula = 𝐀𝐮𝟐 𝐎𝟑 Molecular formula: Like our facebook page: facebook.com/DelzyScholars 12 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 Molecular formula: Formula that shows the actual number of atoms of each element present in a molecule of the compound. (The molecular formula is a whole number multiple of the empirical formula). If the empirical formula and molecular weight/molar mass of a compound are known, the molecular formula of the compound can be determined. The molecular formula is a whole number multiple of the empirical formula; Molecular formula = (Empirical formula)𝑛 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 n = whole number multiple = 𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑤𝑒𝑖𝑔ℎ𝑡 Problem example: A compound has the following composition by mass: C, 0.681 g; H, 0.137 g; O, 0.181 g. (a) Calculate the empirical formula of the compound. (b) If the relative molecular mass of the compound is 88.17, calculate the molecular formula. Solution: C H O Mass (g) 0.681 0.137 0.181 Step-1: 0.681 0.137 0.181 Divide by the atomic mass of the 1 16 12 elements 𝟎. 𝟎𝟓𝟔𝟕 0.137 0.0113 Step-2: 0.0567 0.137 0.0113 Divide through the answers 0.0113 0.0113 0.0113 obtained by the smallest value =1 = 5 = 12 Empirical formula = 𝐂𝟓 𝐇𝟏𝟐 𝐎 (b) Molecular formula = (Empirical formula)𝑛 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠/𝑤𝑒𝑖𝑔ℎ𝑡 𝑛= 𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑤𝑒𝑖𝑔ℎ𝑡 Molecular mass/weight = 88.17g/mol (Will be given or the data to calculate for it given) Empirical formula (𝐂𝟓 𝐇𝟏𝟐 𝐎) Empirical formula weight = (12 × 5) + (1 × 12) + (16 × 1) = 88𝑔/𝑚𝑜𝑙 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠/𝑤𝑒𝑖𝑔ℎ𝑡 88.17 𝑛= = =1 𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑤𝑒𝑖𝑔ℎ𝑡 88 Molecular formula = (Empirical formula)𝑛 Molecular formula = (𝑪𝟓 𝑯𝟏𝟐 𝑶)1 = 𝑪𝟓 𝑯𝟏𝟐 𝑶 Like our facebook page: facebook.com/DelzyScholars 13 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 Calculation of empirical formula via composition by mass from combustion data When a compound containing carbon and hydrogen is completely combusted, the carbon present will be converted to 𝑪𝑶𝟐 , and the hydrogen is converted 𝑯𝟐 𝑶. From the masses of 𝑪𝑶𝟐 and 𝑯𝟐 𝑶 we can calculate the mass of C and H in the original compound and thereby the empirical formula. If a third element (mostly oxygen) is present in the compound, its mass can be determined by subtracting the masses of C and H from the compound’s original mass. Problem example: An organic compound, A, contains only carbon, hydrogen and oxygen. When 1.46g of A, burns in excess oxygen, 2.79g 𝑪𝟎𝟐 and 1.71g 𝑯𝟐 𝑶 are formed. (C =12, H = 1.0) (a) Calculate the empirical formula of A (b) If the relative molecular mass is 92.16, what is the molecular formula of B? 𝑀𝑎𝑠𝑠 𝑜𝑓 𝐶𝑂2 𝑔𝑖𝑣𝑒𝑛 = 2.79𝑔; 𝑀𝑎𝑠𝑠 𝑜𝑓 𝐻2 𝑂 𝑔𝑖𝑣𝑒𝑛 = 1.71𝑔 𝐶 Mass of C = × 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝑂2 given 𝐶𝑂2 12𝑔/𝑚𝑜𝑙  𝐌𝐚𝐬𝐬 𝐨𝐟 𝐂 = × 2.79𝑔 = 𝟎. 𝟕𝟔𝒈 𝑪 44𝑔/𝑚𝑜𝑙 2× 𝐻 Mass of H = × 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐻2 𝑂 𝑔𝑖𝑣𝑒𝑛 𝐻2 𝑂 2 × 1𝑔/𝑚𝑜𝑙  𝐌𝐚𝐬𝐬 𝐨𝐟 𝐇 = × 1.71𝑔 = 𝟎. 𝟏𝟗𝒈 𝑯 18𝑔/𝑚𝑜𝑙 Hint: from the question we were told its contains, C, H & O Check for the mass of oxygen (third element) present, this is obtained by subtracting the mass of both C & H from the original mass of the compound.  Mass of O = 1.46g − (0.76 + 0.19)𝑔 = 𝟎. 𝟓𝟏𝒈 𝑶 C H O Mass (g) 0.76 0.19 0.51 Divide by the atomic mass of 0.76 0.19 0.51 the elements to give no. of mole 12 1 16 𝟎. 𝟎𝟔𝟑 0.19 0.032 Divide by the smallest 0.063 0.19 0.032 0.032 0.032 0.032 = 2 = 6 =1 Empirical formula = 𝑪𝟐 𝑯𝟔 𝑶 (b) Molecular formula = (Empirical formula)𝑛 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠/𝑤𝑒𝑖𝑔ℎ𝑡 92.16 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠/𝑤𝑒𝑖𝑔ℎ𝑡 𝑛= = =2 𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑤𝑒𝑖𝑔ℎ𝑡 46 𝑛= 𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑤𝑒𝑖𝑔ℎ𝑡 Molecular formula = (Empirical formula)𝑛 Molecular mass/weight = 92.16g/mol Molecular formula = (𝑪𝟐 𝑯𝟔 𝑶)𝑛 = (𝑪𝟐 𝑯𝟔 𝑶)2 Empirical formula (𝑪𝟐 𝑯𝟔 𝑶) weight 𝑴𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 = 𝑪𝟒 𝑯𝟏𝟐 𝑶𝟐 (12 × 2) + (1 × 6) + (16 × 1) = 46𝑔/𝑚𝑜𝑙 Like our facebook page: facebook.com/DelzyScholars 14 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 MOLE CONCEPT & STOICHIOMETRY CALCULATION Mole is the amount of matters that contains as many objects as the number of atoms in exactly 12g of 12C. This number is called Avogadro’s constant, it has the symbol of (𝑁𝐴 ), and has the value 𝟔. 𝟎𝟐 × 𝟏𝟎𝟐𝟑. Or simply put; Mole is the fundamental unit in chemistry for measuring “amount of substance” or the “number of particles (atoms, ions, molecules, etc) of substance”. Calculation under mole concept and stoichiometry are further subdivided into two (2) for better understanding and they are as follows:  Relationship between mole and mass  Relatonship between mole, mass and number of particles  RELATIONSHIP BETWEEN MOLE AND MASS: The following formula is used to calculate for mole when mass is known and vice versa 𝑚𝑎𝑠𝑠(𝑔) No of moles = 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑔/𝑚𝑜𝑙 ) Molar mass is the mass in grams of one mole of a substance (molecule or compound), but sometimes the molar mass is still used for showing the mass in one (1) mole of a single atom (in this case it’s called atomic mass). Or simply put: Molar mass is the mass in grams per mole of substance). 1mol of any substance = Molar mass in gram of the substance Molar mass is usually calculated by summing the atomic masses of the various atoms making up the molecules.  Converting mass to no. of mole , millimole etc and vice versa (Mass in 1mole of substance = molar mass [g/mol]) Problem example 1: Calculate the number of moles of glucose (C6H12O6) in 5.380g of glucose. Solution: Mass of glucose given = 5.380g Molar mass of glucose = 180.0g/mol 𝑚𝑎𝑠𝑠(𝑔) No of moles = 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑔/𝑚𝑜𝑙 ) 5.380(𝑔) No of moles of C6 H12 O6 = 180.0 (𝑔/𝑚𝑜𝑙 ) No of moles of C6 H12 O6 in 5.380g glucose = 0.02989 𝐦𝐨𝐥. 𝐂𝟔 𝐇𝟏𝟐 𝐎𝟔 Like our facebook page: facebook.com/DelzyScholars 15 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 Problem example 2: How many millimoles of iron are there in 0.600g of iron? [Fe = 55.85] Solution: Mass of Fe given = 0.600g Molar mass of Fe = 55.85g/mol 𝑚𝑎𝑠𝑠(𝑔) 0.600𝒈 𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 = = = 0.01074𝑚𝑜𝑙 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑔/𝑚𝑜𝑙) 55.85 𝑔/𝑚𝑜𝑙 Convert to millimole (1000mmol = 1mol) 1000mmol 0.01074mol × = 𝟏𝟎. 𝟕𝟒𝐦𝐦𝐨𝐥 1mol Problem example 3: Graphite is the crystalline form of carbon used in “lead” pencils. How many moles of carbon are in 315mg graphite? Solution: First: convert the 315mg (milligram) of C to gram (g) 1g 315mg × = 𝟎. 𝟑𝟏𝟓𝐠 1000mg mass(g) No of moles = molar mass (g/mol ) Molar mass of carbon = 12g/mol 0.315g mole of C = = 0.02625mol 12g/mol Therefore mol of C in 315mg (0.315g) graphite = 0.02625mol Problem example 4: Calculate the number of moles of magnesium (Mg) atoms in 10.0g of magnesium. Solution: Mass of Mg given = 10.0g Molar mass of Mg = 24g/mol 𝑚𝑎𝑠𝑠(𝑔) 10.0 𝒈 𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 = = = 0.42𝑚𝑜𝑙 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑔/𝑚𝑜𝑙) 24 𝑔/𝑚𝑜𝑙 Problem example 5: Calculate the mass of 0.380mol 𝐶𝐻3 𝐶𝑂𝑂𝐻 Solution: 𝑚𝑎𝑠𝑠(𝑔) 𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 = 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑔/𝑚𝑜𝑙) 𝑚𝑎𝑠𝑠(𝑔) = 𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 × 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑔/𝑚𝑜𝑙) Molar mass of 𝐶𝐻3 𝐶𝑂𝑂𝐻 = (12 × 2) + (16 × 2) + (1 × 4) = 60𝑔/𝑚𝑜𝑙 𝑚𝑎𝑠𝑠(𝑔) = 0.380𝑚𝑜𝑙 × 60𝑔/𝑚𝑜𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 0.38𝑚𝑜𝑙 𝐶𝐻3 𝐶𝑂𝑂𝐻 = 22.8𝑔 Like our facebook page: facebook.com/DelzyScholars 16 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576  RELATIONSHIP BETWEEN MOLE, MASS AND NUMBER OF PARTICLES & AVOGADRO’S NUMBER Mole is the amount of matters that contains as many objects as the number of atoms in exactly 12g of 12C. This number is called Avogadro’s constant, it has the symbol of (𝑁𝐴 ), and has the value 𝟔. 𝟎𝟐 × 𝟏𝟎𝟐𝟑. In real sense, the mole is simply a counting number. Just as a Dozen = 12 Experimentally; 1 Mole = 6.023 × 1023 This value 6.023 ×1023 is called Avogadro’s number with symbol NA. If we can say; 1dozen of spoon = 12 spoons Therefore: - 1mole of particles = 6.023 × 1023 number of particles - 1mole of atom = 6.023 × 1023 number of atoms [Example: 1mole of C = 6.023 × 1023 atoms of C] - 1mole of substance = 6.023 × 1023 number of substance - I mole of molecule = 6.023 × 1023 number of molecules  CALCULATION OF NUMBER OF PARTICLE WHEN NUMBER OF MOLE IS GIVEN AND VICE VERSA. The formula relating no of mole and number of particle is as follows: 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝑁𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 = 𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜′ 𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 (𝑁𝐴 ) 𝑁𝐴 is a constant =6.023 × 1023number) Note: The particle can be an atom, ions or molecules. Problem example 6: Graphite is the crystalline form of carbon used in “lead” pencils. How many carbon atoms are in 0.02625mol graphite? [(Avogadro’s number = 6.023×1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙𝑒] Solution: 𝐍𝐨𝐭𝐞: Graphite = carbon, because graphite is allotrope of carbon No of mole of carbon (graphite) = 0.02625mol; 𝑁𝐴 = 6.023 × 1023 ; Number of particles (No of carbon atoms) =? ? Using: 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆 𝑵𝒐. 𝒐𝒇 𝒎𝒐𝒍𝒆 = 𝑨𝒗𝒐𝒈𝒂𝒅𝒓𝒐′ 𝒔 𝒏𝒖𝒎𝒃𝒆𝒓 (𝑵𝑨 ) 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 = 𝑁𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒 × 𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜′ 𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 (𝑁𝐴 ) 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 = 0.02625mol × 6.023 × 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙 = 𝟏. 𝟓𝟖 × 𝟏𝟎𝟐𝟐 𝒂𝒕𝒐𝒎𝒔 𝒐𝒇 𝑪 OR (Can be solved alternatively as) Like our facebook page: facebook.com/DelzyScholars 17 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 We know that: 1mol of C = 6.023×𝟏𝟎𝟐𝟑 atoms of C So therefore; 0.02625mol of C = 𝑿 𝑎𝑡𝑜𝑚𝑠 𝐶 Making X subject of the formula 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎𝟐𝟑 𝒂𝒕𝒐𝒎𝒔 𝑪 × 𝟎. 𝟎𝟐𝟔𝟐𝟓 𝒎𝒐𝒍 𝑪 𝒙= 𝟏 𝒎𝒐𝒍 𝑪 x = 𝟏. 𝟓𝟖 × 𝟏𝟎𝟐𝟐 atoms of C Therefore; 𝟎. 𝟎𝟐𝟔𝟐𝟓 𝒎𝒐𝒍 𝑪 contains 𝟏. 𝟓𝟖 × 𝟏𝟎𝟐𝟐 𝒂𝒕𝒐𝒎𝒔 𝑪 Problem example 5: 1. Calculate the number of molecules in 0.350mol C6H12O6. Solution Here, we are calculating the number of molecules of C6 H12 O6 present in 𝟎. 𝟑𝟓𝟎𝒎𝒐𝒍 C6 𝐻12 𝑂6 No of mole of C6 H12 O6 = 0.350mol; NA = 6.023 × 1023 Number of particles (No of molecules of C6 H12 O6 ) =? ? Using: 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆 𝑵𝒐. 𝒐𝒇 𝒎𝒐𝒍𝒆 = 𝑨𝒗𝒐𝒈𝒂𝒅𝒓𝒐′ 𝒔 𝒏𝒖𝒎𝒃𝒆𝒓 (𝑵𝑨 ) 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 = 𝑁𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒 × 𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜′ 𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 (𝑁𝐴 ) 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 = 0.350mol × 6.023 × 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠/𝑚𝑜𝑙 = 𝟐. 𝟏𝟎𝟖 × 𝟏𝟎𝟐𝟑 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑪𝟔 𝑯𝟏𝟐 𝑶𝟔 OR (Can be solved alternatively as) We know that; 𝟏 𝒎𝒐𝒍 𝐶6 𝐻12 𝑂6 = 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎𝟐𝟑 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶6 𝐻12 𝑂6 Therefore; 𝟎. 𝟑𝟓𝟎 𝒎𝒐𝒍 𝐶6 𝐻12𝑂6 = 𝑿 𝑎𝑡𝑜𝑚𝑠 𝐶6 𝐻12 𝑂6 Making X subject of the formula 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎𝟐𝟑 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑪𝟔 𝑯𝟏𝟐 𝑶𝟔 × 𝟎. 𝟑𝟓𝟎𝒎𝒐𝒍 𝑪𝟔 𝑯𝟏𝟐 𝑶𝟔 𝒙= 𝟏 𝒎𝒐𝒍 𝑪𝟔 𝑯𝟏𝟐 𝑶𝟔 x = 𝟐. 𝟏𝟎𝟖 × 𝟏𝟎𝟐𝟑 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑪𝟔 𝑯𝟏𝟐 𝑶𝟔 Therefore; 𝟎. 𝟑𝟓𝟎 𝐦𝐨𝐥 𝐂𝟔 𝐇𝟏𝟐 𝐎𝟔 contains 𝟐. 𝟏𝟎𝟖 × 𝟏𝟎𝟐𝟑 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑪𝟔 𝑯𝟏𝟐 𝑶𝟔 Problem example 6: Calculate the number of H atoms in 0.350mol C6H12O6. Solution Calculating for the number of H atoms present in 0.350mol C6H12O6. It can be seen that 12H atoms are present in C6H12O6 (𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒; 𝟏 𝒎𝒐𝒍𝒆 𝐶6 𝐻12 𝑂6 = 𝟏𝟐𝑯 𝑎𝑡𝑜𝑚𝑠) If; 𝟏 𝒎𝒐𝒍 𝐶6 𝐻12 𝑂6 = 𝟏𝟐𝑯 𝑎𝑡𝑜𝑚𝑠 × 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎𝟐𝟑 So therefore; 𝟎. 𝟑𝟓𝟎 𝒎𝒐𝒍 𝐶6 𝐻12 𝑂6 = (𝒙) 𝑎𝑡𝑜𝑚𝑠 Like our facebook page: facebook.com/DelzyScholars 18 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 Making X subject of the formula 𝟏𝟐𝑯 𝑎𝑡𝑜𝑚𝑠 × 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎𝟐𝟑 × 𝟎. 𝟑𝟓𝟎𝒎𝒐𝒍 𝑪𝟔 𝑯𝟏𝟐 𝑶𝟔 𝒙= 𝟏 𝒎𝒐𝒍 𝑪𝟔 𝑯𝟏𝟐 𝑶𝟔 𝟐𝟒 x = 𝟐. 𝟓𝟑 × 𝟏𝟎 𝑯 atoms Therefore; 𝟎. 𝟑𝟓𝟎 𝒎𝒐𝒍 𝐶6 𝐻12 𝑂6 contains 𝟐. 𝟓𝟑 × 𝟏𝟎𝟐𝟒 𝑯 𝒂𝒕𝒐𝒎𝒔 Problem example 7:. How many oxygen atoms are in 0.25 mol (𝐶𝑎(𝑁𝑂3 )2 Solution: Calculating for number of oxygen atoms present in 0.25mol 𝑪𝒂(𝑵𝑶𝟑 )𝟐. It can be seen that 6H atoms are present in 𝑪𝒂(𝑵𝑶𝟑 )𝟐 ; (𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒: 𝟏 𝒎𝒐𝒍𝒆 𝑪𝒂(𝑵𝑶𝟑 )𝟐 = 𝟔𝑯 𝑎𝑡𝑜𝑚𝑠) If: 𝟏 𝒎𝒐𝒍 𝑪𝒂(𝑵𝑶𝟑 )𝟐 = 𝟔𝑯 𝑎𝑡𝑜𝑚𝑠 × 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎𝟐𝟑 So therefore: 𝟎. 𝟐𝟓𝒎𝒐𝒍 𝑪𝒂(𝑵𝑶𝟑 )𝟐 = (𝒙) 𝑎𝑡𝑜𝑚𝑠 Making X subject of the formula 6H atoms × 6.023 × 1023 × 0.25mol Ca(NO3 )2 𝑥= 1 mol Ca(NO3 )2 𝟐𝟑 x = 𝟗. 𝟎𝟑 × 𝟏𝟎 𝑶𝒙𝒚𝒈𝒆𝒏 atoms Therefore; 0.25mol 𝐂𝐚(𝐍𝐎𝟑 )𝟐 contains 𝟗. 𝟎𝟑 × 𝟏𝟎𝟐𝟑 𝐎𝐱𝐲𝐠𝐞𝐧 𝐚𝐭𝐨𝐦𝐬 Problem example 8: How many atoms of hydrogen are present in 0.0138mol of ammonia? Solution Calculating number of H atoms present in 0.0138mol ammonia (NH3). It can be seen that 3H atoms are present in NH3 ; (𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒; 𝟏 𝒎𝒐𝒍𝒆 𝑵𝑯𝟑 = 𝟑𝑯 𝑎𝑡𝑜𝑚𝑠) If; 𝟏 𝒎𝒐𝒍 𝑵𝑯𝟑 = 𝟑𝑯 𝑎𝑡𝑜𝑚𝑠 × 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎𝟐𝟑 So therefore; 𝟎. 𝟎𝟏𝟑𝟖 𝒎𝒐𝒍 𝐶6 𝐻12 𝑂6 = (𝒙) 𝑎𝑡𝑜𝑚𝑠 Making X subject of the formula 3𝐻 𝑎𝑡𝑜𝑚𝑠 × 6.023 × 1023 × 0.0138 𝑚𝑜𝑙 𝑁𝐻3 𝑥= 1 𝑚𝑜𝑙 𝐶6 𝐻12 𝑂6 x = 𝟐. 𝟒𝟗 × 𝟏𝟎𝟐𝟐 𝑯 atoms Therefore; 𝟎. 𝟎𝟏𝟑𝟖 𝒎𝒐𝒍 𝑁𝐻3 contains 𝟐. 𝟒𝟗 × 𝟏𝟎𝟐𝟐 𝑯 𝒂𝒕𝒐𝒎𝒔  CALCULATION OF NUMBER OF PARTICLE WHEN MASS IS GIVEN. Problem example 9: How many glucose molecules are in 5.23g of glucose 𝐶6 𝐻12 𝑂6? Solution: Like our facebook page: facebook.com/DelzyScholars 19 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 First: Convert the mass to mole: 𝑚𝑎𝑠𝑠(𝑔) 𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 = 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑔/𝑚𝑜𝑙) Mass given = 5.23g: Molar mass of 𝐶6 𝐻12 𝑂6 = 180g 5.23𝑔 𝑁𝑜. 𝑜𝑓𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 = = 0.0291 𝑚𝑜𝑙 180𝑔/𝑚𝑜𝑙 Second: Solve for the number of particle (number of glucose) using the mole 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆 𝑵𝒐. 𝒐𝒇 𝒎𝒐𝒍𝒆 = 𝑨𝒗𝒐𝒈𝒂𝒅𝒓𝒐′ 𝒔 𝒏𝒖𝒎𝒃𝒆𝒓 (𝑵𝑨 ) 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 = 𝑁𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒 × 𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜′ 𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 (𝑁𝐴 ) 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 = 0.0291 𝑚𝑜𝑙 × 6.023 × 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠/𝑚𝑜𝑙 = 𝟏. 𝟕𝟓 × 𝟏𝟎𝟐𝟐 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 glucose OR (Alternatively can be solved as) Here, we are calculating number of glucose molecules present in 5.23g glucose 𝐂𝟔 𝐇𝟏𝟐 𝐎𝟔. We know that: 𝟏 𝒎𝒐𝒍 𝐶6 𝐻12 𝑂6 = 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎𝟐𝟑 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶6 𝐻12 𝑂6 Note: [𝟏 𝒎𝒐𝒍 𝐶6 𝐻12 𝑂6 = 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶6 𝐻12 𝑂6 ] [Molar mass of 𝐶6 𝐻12 𝑂6 = 180g]  `𝟏𝟖𝟎𝐠 𝐶6 𝐻12 𝑂6 = 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎𝟐𝟑 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶6 𝐻12 𝑂6 Therefore; 𝟓. 𝟐𝟑𝐠 𝐶6 𝐻12 𝑂6 = 𝑿 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶6 𝐻12 𝑂6 Making X subject of the formula 𝟓. 𝟐𝟑𝐠 C6 H12 O6 × 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎𝟐𝟑 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 C6 H12 O6 𝒙= 𝟏𝟖𝟎𝐠 𝐶6 𝐻12 𝑂6 x = 𝟏. 𝟕𝟓 × 𝟏𝟎𝟐𝟐 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 C6 H12 O6  CALCULATION OF MASS WHEN NUMBER OF PARTICLES ARE GIVEN. Problem example 10: Calculate the mass contained in each of the following (a) 2.8 × 1024 𝑎𝑡𝑜𝑚𝑠 𝐹𝑒 (b) 6.45 × 1022 𝑎𝑡𝑜𝑚𝑠 𝐶 (c) 1.94 × 1020 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶𝑙2 (d) 8.2 × 1021 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶𝑂2 Solution: (We know: 1mole =6.023 × 1023number) (Recall; 1mol = molar mass) So therefore; (Molar mass = 6.023 × 1023number) Like our facebook page: facebook.com/DelzyScholars 20 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 (a): Molar mass of Fe = 55.85g 6.023 × 1023 atoms of Fe = 55.85g Fe So therefore: 𝟐. 𝟖 × 𝟏𝟎𝟐𝟒 𝒂𝒕𝒐𝒎𝒔 𝐹𝑒 = x g Fe Making x subject of the formula 𝟐. 𝟖 × 𝟏𝟎𝟐𝟒 𝒂𝒕𝒐𝒎𝒔 𝐹𝑒 × 𝟓𝟓. 𝟖𝟓𝐠 Fe 𝑥= 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎𝟐𝟑 𝒂𝒕𝒐𝒎𝒔 𝐹𝑒 x= 259.6g Fe (b): Molar mass of C = 12g 6.023 × 1023 atoms C = 12g C So therefore: 𝟔. 𝟒𝟓 × 𝟏𝟎𝟐𝟐 𝒂𝒕𝒐𝒎𝒔 C = x g Fe Making x subject of the formula 𝟔. 𝟒𝟓 × 𝟏𝟎𝟐𝟐 𝒂𝒕𝒐𝒎𝒔 C × 𝟏𝟐𝐠 C 𝑥= 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎𝟐𝟑 𝒂𝒕𝒐𝒎𝒔 C x= 259.6g Fe (c): Molar mass of C𝑙2 = 71g; [Cl + Cl] = 35.5 + 35.5 =71g 6.023 × 1023 molecules C𝑙2 = 71g C𝑙2 So therefore: 1.94 × 1024 molecules C𝑙2 = x g C𝑙2 Making x subject of the formula 𝟏. 𝟗𝟒 × 𝟏𝟎𝟐𝟒 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆𝒔 C𝑙2 × 𝟕𝟏𝐠 C𝑙2 𝑥= = 228.7𝑔 𝐶𝑙2 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎𝟐𝟑 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆𝒔 C𝑙2 Problem example 11: How many atoms of hydrogen are present in 𝟎. 𝟐𝟑𝟓𝐠 ammonia NH3 ? Solution Here, we are calculating number of H atoms present in 0.235g ammonia (NH3). It can be seen that 3H atoms are present in NH3 ; (𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒; 𝟏 𝒎𝒐𝒍𝒆 𝑵𝑯𝟑 = 𝟑𝑯 𝑎𝑡𝑜𝑚𝑠) [1mol NH3 = molar mass of NH3: Molar mass of NH3 = 17g] If; 17g NH3 = 𝟑𝑯 𝑎𝑡𝑜𝑚𝑠 × 𝟔. 𝟎𝟐𝟑 × 𝟏𝟎𝟐𝟑 So therefore; 𝟎. 𝟐𝟑𝟓𝐠 NH3 = (𝒙) 𝑎𝑡𝑜𝑚𝑠 Making X subject of the formula 3𝐻 𝑎𝑡𝑜𝑚𝑠 × 6.023 × 1023 × 𝟎. 𝟐𝟑𝟓𝐠 NH3 𝑥= 17g NH3 x = 𝟐. 𝟒𝟗 × 𝟏𝟎𝟐𝟐 𝑯 atoms Therefore; 𝟎. 𝟐𝟑𝟓𝐠 𝑁𝐻3 contains 𝟐. 𝟒𝟗 × 𝟏𝟎𝟐𝟐 𝑯 𝒂𝒕𝒐𝒎𝒔 Like our facebook page: facebook.com/DelzyScholars 21 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 BALANCING OF CHEMICAL EQUATION Algebraic method: This is a systematic way of balancing reaction, several rules has been set aside to help guide and aid easy balancing of complex chemical equation using the algebraic method and the include: There are various approach of balancing chemical equation using algebraic method, But fr this material we would be discussing approach-1 and approach-2.  APPROACH - 1: Rules for balancing chemical equation using algebraic method  Rule-1: choose one of the reactant molecule with least number of atom and assign it coefficient to one (1); in a case of decomposition reaction where one reactant decomposes to give 2 or more products, choose the molecule of the product with the least atoms and assign its coefficient to one(1). NOTE: Once one of the reactants has been assigned a coefficient of one (1), nothing else is expected to be assigned to the reactant.  Rule-2: check atoms that occurs once on both sides of the reactant and products and balance their coefficient with same alphabetic coefficient. Once we mean appearing only in one of the reactant and product irrespective of the atomicity.  Rule-3: assign a separate alphabet to the remaining atom that occurs twice on both sides.  Rule-4: balance the equation by solving for the alphabet and replacing them into the equation. For the purpose of digesting the rules above, the following example would be cited. Problem example: Balance the following equation 𝑆𝑏2 𝑂3 + 𝐶 → 𝑆𝑏 + 𝐶𝑂2 Solution: Rule-1: choose the reactant with the least number of atoms and assign it coefficient to one (1); Carbon is just one atom and as such would be assigned 1. 𝑆𝑏2 𝑂3 + 𝟏𝐶 → 𝑆𝑏 + 𝐶𝑂2 Rule-2: check atoms that occurs once on both sides; balance such atoms using the same alphabet. From the equation Sb & O occurs once on both side;  Sb occured only in 𝑺𝒃𝟐 𝑶𝟑 on the reactant and Sb on the product.  O occured only in 𝑆𝑏2 𝑂3 on the reactant and CO2 on the product. 𝟑𝒂 𝒂 𝑆𝑏2 𝑂3 + 𝟏𝐶 → 𝟐𝒂 𝑆𝑏 + 𝐶𝑂2 𝟐 Rules-3 would be skipped since alphabet a has covered for all atoms Rule-4: solving for the alphabet and replacing them into the equation. Sb: 2a = 2a [balanced] Like our facebook page: facebook.com/DelzyScholars 22 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 3𝑎 O: 3a = × 2  3𝑎 = 3𝑎 [𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑] 2 𝟑𝒂 𝟐 C: 1=  2 = 3a;  a= 𝟐 𝟑 𝟐 Replacing a with in the equation 𝟑 𝟐 𝟐 3 𝟐 𝑆𝑏2 𝑂3 + 𝐶 → 2(𝟑) 𝑆𝑏 + ( ) 𝐶𝑂2 𝟑 2 𝟑 𝟐 𝟒 𝑆𝑏2 𝑂3 + 𝐶 → 𝑆𝑏 + 𝐶𝑂2 𝟑 𝟑 Multiplying through by 3 so as to remove the fraction 𝟐𝑺𝒃𝟐 𝑶𝟑 + 𝟑𝑪 → 𝟒𝑺𝒃 + 𝟑𝑪𝑶𝟐 [𝒃𝒂𝒍𝒂𝒏𝒄𝒆𝒅 ] Problem example: Balance the equation; 𝑲𝟐 𝑪𝑶𝟑 + 𝑨𝒍𝟐 𝑪𝒍𝟔 → 𝑨𝒍𝟐 (𝑪𝑶𝟑 )𝟑 + 𝑲𝑪𝒍 Solution: Rule-1: choose one of the reactant with least atoms and assign coefficient of one (1). {𝑲𝟐 𝑪𝑶𝟑 = 𝟏} 〈𝟏〉𝐾𝟐 𝐶𝑂3 + 𝐴𝑙2 𝐶𝑙6 → 𝐴𝑙2 (𝐶𝑂3 )3 + 𝐾𝐶𝑙 Rule-2: check for atoms that occur once on both side and balance with same alphabet; [K, C, O, Al & Cl all are occurring once] 〈𝟏〉𝑲𝟐 𝑪𝑶𝟑 + 〈𝒂〉𝑨𝒍𝟐 𝑪𝒍𝟔 → 〈𝒂〉𝑨𝒍𝟐 (𝑪𝑶𝟑 )𝟑 + 〈𝟔𝒂〉 𝑲𝑪𝒍 We use (a) to balance Al; but since the (a) added to Al2Cl6 will definitely affect Cl, we use it as well to balance Cl since Cl is also occurring once. So rule-3 will be skipped. Rule-4: solving for the alphabet and replacing them into the equation. Al: 2a = 2a [balanced] Cl: 6a = 6a [balanced] 2 1 K: 2 = 6a  a = = 6 3 𝟏 Therefore; a=𝟑 𝟏 𝟏 𝟏 〈𝟏〉𝑲𝟐 𝑪𝑶𝟑 + 〈 〉 𝑨𝒍𝟐 𝑪𝒍𝟔 → 〈 〉 𝑨𝒍𝟐 (𝑪𝑶𝟑 )𝟑 + 〈𝟔 × 〉 𝑲𝑪𝒍 𝟑 𝟑 𝟑 Multiplying through by 3 so as to remove the fraction 〈𝟑〉𝑲𝟐 𝑪𝑶𝟑 + 𝑨𝒍𝟐 𝑪𝒍𝟔 → 𝑨𝒍𝟐 (𝑪𝑶𝟑 )𝟑 + 〈𝟔〉 𝑲𝑪𝒍 Like our facebook page: facebook.com/DelzyScholars 23 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576  APPROACH – 2 Steps involved here:  Step - 1: Insert/assign an unknown coefficient in front of each molecular species in the equation.  Step - 2: Write a balanced equation for each element/atom interms of the unknown coefficient  Step - 3: Choose one of the unknown coefficent and assign its value as one (1). The choice of which coefficient to set to one (1) is completely arbitrary and may be taken to be the one that is most convenient for solving the problem. Problem example: Balance the equation; 𝐶𝑎3 (𝑃𝑂4 )2 + 𝐻2 𝑆𝑂4 → 𝐶𝑎(𝐻2 𝑃𝑂4 )2 + 𝐶𝑎(𝐻𝑆𝑂4 )2 Solution: Step – 1: Assign an unkown coefficient in front of each molecular species in the equation. 〈𝐚〉Ca3 (PO4 )2 + 〈𝐛〉H2 SO4 → 〈𝐜〉Ca(H2 PO4 )2 + 〈𝐝〉Ca(HSO4 )2 Step - 2: 𝑪𝒂: 3𝑎 = 𝑐 + 𝑑 … … … …. (𝟏) Write a balanced equation for 𝐏: 2𝑎 = 2𝑐 … … ….. (𝟐) each element/atom interms of 𝐎: 8𝑎 + 4𝑏 = 8𝑐 + 8𝑑 ….. (𝟑) the unknown coefficient. 𝐇: 2𝑏 = 4𝑐 + 2𝑑 ….. (𝟒) 𝐒: 𝑏 = 2𝑑 …. ….. (𝟓) Step – 3: 𝑙𝑒𝑡 𝒂 = 𝟏 Assign one of the coefficient From eqn(𝟐): 𝒂 = 𝒄 ∶ 𝒂 = 𝟏, 𝒄 = 𝟏 of value of 1; preferably the From eqn(𝟏): 𝟑𝒂 = 𝒄 + 𝒅 one that’s more convenient for 3 = 1 + 𝑑 → 3 − 1 = 𝑑: 𝒅 =𝟐 solving for other unknowns. From eqn(𝟓): 𝐛 = 𝟐𝒅: 𝒃 = 𝟐(𝟐) = 𝟒 𝒂 = 𝟏, 𝒃 = 𝟒, 𝒄 = 𝟏, 𝒅=𝟐 〈𝐚〉Ca3 (PO4 )2 + 〈𝐛〉H2 SO4 → 〈𝐜〉Ca(H2 PO4 )2 + 〈𝐝〉Ca(HSO4 )2 𝐂𝐚𝟑 (𝐏𝐎𝟒 )𝟐 + 𝟒𝐇𝟐 𝐒𝐎𝟒 → 𝐂𝐚(𝐇𝟐 𝐏𝐎𝟒 )𝟐 + 𝟐𝐂𝐚(𝐇𝐒𝐎𝟒 )𝟐 Like our facebook page: facebook.com/DelzyScholars 24 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 Problem example: Balance the equation: 𝑲𝟐 𝑪𝑶𝟑 + 𝑨𝒍𝟐 𝑪𝒍𝟔 → 𝑨𝒍𝟐 (𝑪𝑶𝟑 )𝟑 + 𝑲𝑪𝒍 Solution: Step – 1: Assign an unkown coefficient in front of each molecular species in the equation. 〈𝒂〉𝐊 𝟐 𝐂𝐎𝟑 + 〈𝐛〉𝐀𝐥𝟐 𝐂𝐥𝟔 → 〈𝐜〉𝐀𝐥𝟐 (𝐂𝐎𝟑 )𝟑 + 〈𝐝〉 𝐊𝐂𝐥 Step - 2: 𝑲: 2𝑎 = 𝑑 … … … …. (𝟏) Write a balanced equation for 𝐂: 𝑎 = 3𝑐. … … ….. (𝟐) each element/atom interms of 𝐎: 3𝑎 = 9𝑐 … ….. ….. (𝟑) the unknown coefficient. 𝐀𝐥: 2𝑏 = 2𝑐 … … … ….. (𝟒) 𝐂𝐥: 𝟔𝑏 = 𝑑 … … …. ….. (𝟓) Step – 3: 𝑙𝑒𝑡 𝒂 = 𝟏 Assign one of the coefficient From eqn(𝟏): 2𝑎 = 𝑑 ∶ 𝐝 = 𝟐(𝟏) = 𝟐 of value of 1;jii preferably the 𝟏 one that’s more convenient for From eqn(𝟐): 𝑎 = 3𝑐: 𝟏 = 𝟑𝒄: 𝒄 = 𝟑 solving for other unknowns. 𝟏 From eqn(𝟒): 2𝑏 = 2𝑐: 𝑏 = 𝑐: 𝒃 = 𝟑 𝟏 𝟏 𝒂 = 𝟏, 𝒃= , 𝒄= , 𝒅=𝟐 𝟑 𝟑 〈𝐚〉K 2 CO3 + 〈𝐛〉Al2 Cl6 → 〈𝐜〉Al2 (CO3 )3 + 〈𝐝〉 KCl 𝟏 𝟏 K 2 CO3 + Al2 Cl6 → Al2 (CO3 )3 + 𝟐 KCl 𝟑 𝟑 Multiply through by 3: 𝟑𝐊 𝟐 𝐂𝐎𝟑 + 𝐀𝐥𝟐 𝐂𝐥𝟔 → 𝐀𝐥𝟐 (𝐂𝐎𝟑 )𝟑 + 𝟔 𝐊𝐂𝐥 Problem example: Balance the equation: 𝐒𝐛𝟐 𝐎𝟑 + 𝐂 → 𝐒𝐛 + 𝐂𝐎𝟐 Solution: Step – 1: Assign an unkown coefficient in front of each molecular species in the equation. 〈𝒂〉Sb2 O3 + 〈𝐛〉 C → 〈𝐜〉Sb + 〈𝐝〉 CO2 Step - 2: 𝐒𝐛: 2𝑎 = 𝑐 … … … …. (𝟏) Write a balanced equation for 𝐎: 3𝑎 = 2𝑑.. … … ….. (𝟐) each element/atom interms of 𝐂: 𝑏 = 𝑑. … … ….. (𝟑) the unknown coefficient. Step – 3: 𝑙𝑒𝑡 𝒃 = 𝟏 Assign one of the coefficient From eqn(𝟑): 𝑏 = 𝑑 ∶ 𝒃 = 𝟏, 𝒅 = 𝟏 of value of 1; preferably the From eqn(𝟐): 3𝑎 = 2𝑑: 3𝑎 = 2(1) one that’s more convenient for 𝟐 𝒂= 𝟑 solving for other unknowns. From eqn(𝟏): 2𝑎 = 𝑐: 𝟐 𝟒 𝒄 = 𝟐𝒂 = 𝟐 ( 𝟑) = 𝟑 Like our facebook page: facebook.com/DelzyScholars 25 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 𝟐 𝟒 𝒂= , 𝒃 = 𝟏, 𝒄= , 𝒅=𝟏 𝟑 𝟑 〈𝒂〉Sb2 O3 + 〈𝐛〉 C → 〈𝐜〉Sb + 〈𝐝〉 CO2 𝟐 𝟒 Sb2 O3 + C → Sb + CO2 𝟑 𝟑 Multiply through by 3 𝟐𝐒𝐛𝟐 𝐎𝟑 + 𝟑𝐂 → 𝟒𝐒𝐛 + 𝟑𝐂𝐎𝟐 Problem example: Balance the equation; 𝐇𝟐 𝐎 + 𝐊𝐎𝟐 → 𝐊𝐎𝐇 + 𝐎𝟐 Solution: Step – 1: Assign an unkown coefficient in front of each molecular species in the equation. 〈𝒂〉 〈𝒂〉H2 O + 〈𝒃〉KO2 → 〈𝐜〉KOH + 〈𝐝〉O2 Step - 2: 𝐇: 2𝑎 = 𝑐 … … … …. (𝟏) Write a balanced equation for 𝐎: 𝑎 + 2𝑏 = 𝑐 + 2𝑑... ….. (𝟐) each element/atom interms of 𝐊: 𝑏 = 𝑐. ….. … ….. (𝟑) the unknown coefficient. Step – 3: 𝑙𝑒𝑡 𝒃 = 𝟏 Assign one of the coefficient From eqn(𝟑): 𝑏 = 𝑐 ∶ 𝒃 = 𝟏, 𝒄 = 𝟏 of value of 1; preferably the From eqn(𝟏): 2𝑎 = 𝑐: 𝟐𝒂 = 𝟏 one that’s more convenient for 𝟏 𝒂= 𝟐 solving for other unknowns. 𝟏 From eqn(𝟐): 𝑎 + 2𝑏 = 𝑐 + 2𝑑: 𝟐 + 2(1) = 1 1 + 2𝑑: + 2 − 1 = 2𝑑 2 3 3 𝟑 = 2𝑑: 2𝑑 = : 𝒅=𝟒 2 2 𝟏 𝟑 𝒂= , 𝒃 = 𝟏, 𝒄 = 𝟏, 𝒅= 𝟐 𝟒 〈𝒂〉H2 O + 〈𝒃〉KO2 → 〈𝐜〉KOH + 〈𝐝〉O2 1 3 H2 O + KO2 → KOH + O2 2 4 Multiply through by 4 𝟐𝐇𝟐 𝐎 + 𝟒𝐊𝐎𝟐 → 𝟒𝐊𝐎𝐇 + 𝟑𝐎𝟐 Like our facebook page: facebook.com/DelzyScholars 26 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 CALCULATION INVOLVING REACTING MASS This calculation involves the use of a balanced chemical equation to derive the stoichiometric factors relating amounts of reactants and products A balanced chemical equation provides a great deal of information in a very succinct format. The coefficient of the reactants and products species provide the relative numbers of these chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. These quantitative relationships are known as the reaction’s stoichiometry, a term derived from the Greek words stoicheion (meaning “element”) and metron (meaning “measure”). Sample Problem: What mass (g) of iron (Fe) will be produced from 152.6 g of carbon monoxide and an excess of iron (III) oxide? [Fe = 55.85g/mol; C = 12g/mol; O = 16g/mol;] 𝐹𝑒2 𝑂3(𝑠) + 3𝐶𝑂(𝑔) → 2𝐹𝑒(𝑠) + 3𝐶𝑂2(𝑔) Solution: 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐹𝑒 = 55.85𝑔/𝑚𝑜𝑙: 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶𝑂 = 12 + 16 = 28𝑔/𝑚𝑜𝑙 From the equation: 3 𝑚𝑜𝑙 𝐶𝑂 → 2 𝑚𝑜𝑙 𝐹𝑒 3 × 28𝑔 𝑪𝑶 → 2 × 55.85𝑔 𝑭𝒆 152.6𝑔 𝑪𝑶 → 𝑥𝑔 𝑭𝒆 𝑐𝑟𝑜𝑠𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑎𝑛𝑑 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑥 2 ×55.85𝑔 𝐹𝑒 × 152.6𝑔 𝐶𝑂 𝑥= = 202.9𝑔 𝐹𝑒 3 × 28𝑔 𝐶𝑂 Sample Problem: How many grams of 𝐶𝑂2 are produced in the combustion of 50.0 g of propane, 𝐶3 𝐻8 ? The molar mass of 𝐶𝑂2 is 44.0 g/mol and the molar mass of propane is 44.0 g/mol. 𝐶3 𝐻8(𝑔) + 𝑂2(𝑔) → 3𝐶𝑂2(𝑔) + 𝐻2 𝑂(𝑙) Solution: From the equation: 1 𝑚𝑜𝑙 𝐶3 𝐻8 → 3 𝑚𝑜𝑙 𝐶𝑂2 44𝑔 𝐶3 𝐻8 → 3 × 44.0𝑔 𝐶𝑂2 50.0𝑔 𝐶3 𝐻8 → 𝑥𝑔 𝐶𝑂2 𝑐𝑟𝑜𝑠𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑎𝑛𝑑 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑥 3×44.0𝑔 𝐶𝑂2 × 50.0𝑔 𝐶3 𝐻8 𝑥= = 150𝑔 𝐶𝑂2 44𝑔 𝐶3 𝐻8 Sample Problem: Determine the amount of oxygen produced when 0.549 g of KClO3 decomposes. The molar mass of 𝑂2 is 32.0 g/mol and the molar mass of 𝐾𝐶𝑙𝑂3 is 122.6 g/mol. 𝟐KClO3(s) → 𝟐KCl(s) + 3O2(g) 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: From the equation: 2 𝑚𝑜𝑙 KClO3 → 3 𝑚𝑜𝑙 𝑂2 2 × 122.6𝑔 KClO3 → 3 × 32.0𝑔𝑂2 0.549𝑔 KClO3 → 𝒙 𝑔 𝑂2 < 𝑐𝑟𝑜𝑠𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑎𝑛𝑑 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑥 3×32.0𝑔 𝑂2 × 0.549𝑔 KClO3 𝑥= = 0.215𝑔 𝑂2 2×122.6𝑔 KClO3 Like our facebook page: facebook.com/DelzyScholars 27 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 LIMITING REAGENT/ REACTANT Limiting reagent is the reagent/reactant that is completely consumed during a chemical reaction. Once this reagent/reactant is completely consumed, the reaction will stop and as such there would be no further products. In order words limiting reagent can also be defined as a chemical reactant that limits the amount of product that is formed during a chemical reaction. The maximum amount (or simply; the amount of product) obtained in a chemical reaction when a limiting reagent/reactant is completely consumed is called THEORETICAL YIELD. Those reagents/reactants that are left over once the limiting reagent is completely consumed are called Excess reagent/reactant. Illustration1: when two or more reactants are made to react during a chemical reaction and as usual every chemical reaction must give rise to new products, the first reactant to be completely consumed by the reaction is the limiting reagent/reactant, this would determine the amount of product that would be formed because once the reactant finishes/completely consumed, the reaction would stop. (By delzy) Illustration2: Cars uses combustible engines, these engines makes use of two reagents [FUEL/DIESEL and AIR (oxygen)]. Combustion reaction takes place inside the combustible engine. Once the engine is put on, FUEL and AIR (from the carburetor) would react to produce carbon dioxide and water which would be emitted through the exhaust pipe. FUEL is always the limiting reagent and AIR is always the excess reagent, because fuel is always the one to finish first and once the fuel finishes the car engine would go off and the car if on motion would be brought to a halt gradually. The distance covered by the car would depend on the amount of fuel present, fuel here is the limiting reagent and will limits the amount of product (in this case, distance covered by the car) during chemical reaction (in this case, combustion reaction inside the car engine that lead to the car to move). (By delzy) CALCULATION OF LIMITNG REAGENTS In calculating for limiting reagent/reactant, two or more reactants with their masses would always be given. Any reaction that has a limiting reagent must also have an excess reagent, so in order to proceed with calculation of limiting reagent; you have to ensure two or more reactants with their masses are given. Except if the limiting reagent was given, in this case there would be no need to calculate for it. Note:  Limiting reagent could also be clearly stated or given in the question. (in this case; No calculation is required)  The chemical reaction that involves the limiting reagent must be balanced with the right stoichiometric coefficient before proceeding with calculation. Steps and procedures in calculating limiting reagents 1. Find the moles of each reactant present. Usually the amounts of the reactants are given in grams and as such would have to be converted into mole; 𝐫𝐞𝐚𝐜𝐭𝐢𝐧𝐠 𝐦𝐚𝐬𝐬(𝐠) mole = 𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 (𝐠/𝐦𝐨𝐥) 2. Divide the mole of each reactant obtained from step1 above by their stoichiometric coefficient in the balanced chemical equation 3. The smallest value from step 2 becomes the limiting reagents. That is; the reactant with the smallest value after dividing their mole by their stoichiometric coefficient becomes the limiting reagent. Like our facebook page: facebook.com/DelzyScholars 28 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 Problem example: At high temperature, sulfur combines with iron to form the brown black iron (II) sulfide; 𝑭𝒆 + 𝑺 → 𝑭𝒆𝑺 If 7.62g of the Fe are allowed to react with 8.67g of S, what is the limiting reagent? Solution: Recall: In calculating for limiting reagent/reactant, two or more reactants with their masses would always be given. In this case, two reactants (Fe & S) and their masses are given. Applying the following steps above: 1. Convert the masses of the reactants to mole by dividing by their molar mass. Fe S reacting mass(g) reacting mass(g) mole = mole = molar mass (g/mol) molar mass (g/mol) 7.62g 8.67g mole of Fe = = 𝟎. 𝟏𝟑𝟔𝒎𝒐𝒍 mole of S = = 𝟎. 𝟐𝟕𝟏𝒎𝒐𝒍 56 g/mol 32 g/mol 2. Divide the mole by their stoichiometric coefficient in the balance equation. [𝟏𝑭𝒆 + 𝟏𝑺 → 𝟏𝑭𝒆𝑺 ] Fe S 0.136 0.271 = 𝟎. 𝟏𝟑𝟔 = 0.271 1 1 3. The smallest value is 0.136 for Fe; therefore Fe IS THE LIMITING REAGENT/REACTANT. Problem example: What is the limiting reagent in the reaction of 5.0g of N2 and 5.0g H2?? 𝟑𝐇𝟐 + 𝐍𝟐 → 𝟐𝐍𝐇𝟑 Solution: Recall, Limiting reagent is calculated when at least two (2) reactants and their masses are given. In this case, two reactants (𝐇𝟐 & 𝐍𝟐 ) and their masses are given. Steps in calculating limiting reagent 1. Convert the masses of the reactants to mole by dividing by their molar mass. 𝐇𝟐 𝐍𝟐 reacting mass(g) reacting mass(g) mole = mole = molar mass (g/mol) molar mass (g/mol) 5.0g 5.0g mole of 𝐇𝟐 = = 𝟐. 𝟒𝟗𝒎𝒐𝒍 mole of 𝐍𝟐 = = 𝟎. 𝟏𝟖𝒎𝒐𝒍 2.008 g/mol 28 g/mol 2. Divide the mole by their stoichiometric coefficient in the balance equation. [ 𝟑𝐇𝟐 + 𝐍𝟐 → 𝟐𝐍𝐇𝟑 ] 𝐇𝟐 𝐍𝟐. 𝟎.𝟏𝟖 𝟐. 𝟒𝟗 = 0.83 = 0.18 1 3 3. Therefore; 0.18 is the smallest value and as such N2 IS THE LIMITING REAGENT. Like our facebook page: facebook.com/DelzyScholars 29 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 Application of limiting reagent The major application of limiting reagent is in the determination of theoretical yield.  THEORETICAL YIELD: Theoretical yield is the maximum amounts of product expected when all of the limiting reagents are totally consumed during a chemical reaction. Theoretical yield is always determined by doing a stoichiometry calculation from the balanced chemical equation. Note: You may be ask to calculate for the Amount of product or Maximum amount of product [usually in gram] of a reaction involving a limiting reagent, please bear in mind that it’s the same as theoretical yield. Recall: Once the limiting reagent finishes/totally used up, the reaction would stop and the product obtained after the reaction has stopped is called the THEORETICAL YIELD (also know that this can also be called amount of product or maximum amount of product) Note: In some cases, you may be given the limiting reagent, while in some other, two or more reagents with their masses will be given, you will have to calculate to find the limiting reagent in this case before you can proceed in applying it for the determination of the theoretical yield. Problem example: At high temperature, sulfur combines with iron to form the brown black iron (II) sulfide; 𝑭𝒆 + 𝑺 → 𝑭𝒆𝑺 If 7.62g of the Fe are allowed to react with 8.67g of S, what is the limiting reagent and calculate the mass of FeS formed. Solution: (a) Calculation of limiting reagent: Fe = limiting reagent (solved already in the previous example) (b) calculation of mass of FeS formed (theoretical yield) 𝑭𝒆 + 𝑺 → 𝑭𝒆𝑺 Stoichiometry calculation from the balanced chemical equation. From the equation:  1mol 𝐅𝐞 = 1mol FeS In grams:  1 × 56g 𝐅𝐞 = 1 × 88g 𝐅𝐞𝐒 (1mol= molar mass) So therefore, 7.62g Fe = x g FeS 88g FeS × 7.62g Fe Cross multiply and solve for x; 𝑥 = = 12.0g FeS 56g Fe Therefore the mass of FeS formed (theoretical yield) = 12.0g FeS Note: Just in case; you may be ask to present the answer in mole; simply divide the mass by the molar mass (I said just in case, although it’s rare) Problem example: Take the reaction: 4𝐍𝐇𝟑 + 𝟓𝐎𝟐 → 𝟒𝐍𝐎 + 𝟔𝐇𝟐 𝐎. In an experiment, 3.25g of NH3 are allowed to react with 3.50g of O2. (a) What is the limiting reagent? (b) What is the theoretical yield of NO? Like our facebook page: facebook.com/DelzyScholars 30 | P a g e  DELZY EXPLICIT CHM MATERIAL/GUIDE……. 07038101576 Answer: (a). Calculation of limiting reagent: 𝐍𝐇𝟑 𝐎𝟐 1. Convert the masses of the reactants to 3.25g 3.50g mole by dividing by their atomic masses: 17g/mol 32g/mol 0.191mol 0.109mol 2. Divide the mole by their stoichiometric 0.191 0.109 coefficient in the balance chemical 4 5 equation. = 0.0478 = 0.0218 3. The smallest of the answer obtained 0.0218 is the smallest, from step would give the reactant that is the limiting rea gent 𝐎𝟐 𝒊𝒔 𝒕𝒉𝒆 𝑳𝑰𝑴𝑰𝑻𝑰𝑵𝑮 𝑹𝑬𝑨𝑮𝑬𝑵𝑻 LIMITING REAGENT = 𝐎𝟐 (b). Calculation of Theoretical yield of NO: Molar mass of 𝐎𝟐 = (16 × 2) = 32gmol−1 Molar mass of NO = (14 × 16) = 30gmol−1 Theoretical yield is calculated from the balance chemical equation using the limiting reagent 4𝐍𝐇𝟑 + 𝟓𝐎𝟐 → 𝟒𝐍𝐎 + 𝟔𝐇𝟐 𝐎 < From the equation:  5 mol 𝐎𝟐 = 4 mol 𝐍𝐎 In grams:  5 × 32g 𝐎𝟐 = 4 × 30g 𝐍𝐎 (1mol= molar mass) So therefore, 3.50g 𝐎𝟐 = 𝑥 g 𝐍𝐎 3.50g 𝐎𝟐 × 4×30g 𝐍𝐎 Cross multiply and solve for x; 𝑥 = =2.625g 𝐍𝐎 5 ×32g 𝐎𝟐 Theoretical yield of 𝐍𝐎 = 𝟐. 𝟔𝟐𝟓𝐠 𝐍𝐎 Problem example: An industrial chemist DELZY while in Germany carried out a reaction by reacting 75.0g 𝐏𝟒 with excess chlorine gas to produce 110g 𝐏𝐂𝐥𝟑 in the lab. Given the equation of the reaction below, what is the theoretical yield? 𝐏𝟒 + 𝟔𝐂𝐥𝟐 → 𝟒𝐏𝐂𝐥𝟑 Answer: (a). THEORETICAL YIELD CALCULATION OF 𝐏𝐂𝐥𝟑 Since chlorine (𝐂𝐥𝟐 ) is in excess, it implies the other reactant is the limiting. Therefore Phosphorus (𝐏𝟒 ) is the limiting reagent. Molar mass of 𝐏𝐂𝐥𝟑 = 31 + (35.5 × 3) = 137.5gmol−1 Molar mass of 𝐏𝟒 = (31 × 4) = 124gmol−1 Theoretical yield is calculated from the balanced chemical equation using the limiting reagent 𝐏𝟒 + 𝟔𝐂𝐥𝟐 → 𝟒𝐏𝐂𝐥𝟑 From the equation:  1 mol 𝐏𝟒 = 4 mol 𝐏𝐂𝐥𝟑 In grams:  124g 𝐏𝟒 = 4 × 137.5g 𝐏𝐂𝐥𝟑 (1mol= molar mass) So therefore, 75.0g 𝐏𝟒 = x g 𝐏𝐂𝐥𝟑 75.0g 𝐏𝟒 × 4×137.5g 𝐏𝐂𝐥𝟑 Cross multiply and solve for x; 𝑥 =