General Chemistry (1) - Benha University Past Paper PDF

Benha university Faculty of science Chemistry Department General Chemistry (1) Chapter 1 STOICHIOMETRY Modern chemistry began when chemists recognized the importance of careful measurements and began to ask questions which could be answered quantitatively. Sto...

Benha university Faculty of science Chemistry Department General Chemistry (1) Chapter 1 STOICHIOMETRY Modern chemistry began when chemists recognized the importance of careful measurements and began to ask questions which could be answered quantitatively. Stoichiometry is the branch of chemistry that deals with the quantitative relationships between elements and compounds in chemical reactions. The atomic theory of matter is basic to this study. Dalton's Atomic Theory Many scientists from ancient times believed that all matter consists of atoms, but Dalton made the atomic theory quantitative by showing that it is possible to determine the relative masses of the atoms of different elements. The principle postulates of Dalton theory are: 1- Elements are composed of extremely small particles called atoms. All atoms of the same elements are alike, and atoms of different elements are different 2- The separation and union of atoms occur in chemical reactions. In the reactions, no atom is created or destroyed, and no atom of one element is converted into an atom of another element. 3- A chemical compound is the result of the combination of atoms of two or more elements in a simple numerical ratio. Dalton believed that all atoms of a given element have equal atomic masses. Today we know that a given element may consist of several types 1 of atoms that differ in mass (isotopes). All atoms of the same element, however, react chemically in the same way. The second postulate of Dalton's theory account for the law of conservation of mass which states that "there is no detectable change in mass during the course of a chemical reaction." So, the total of all the materials entering into a chemical reaction must equal the total mass of all the products of the reaction (balanced chemical equation). The third postulate of the theory explains the law of definite proportions which states that "a pure compound always contains the same elements combined by the same proportions by mass." On the basis of his theory, Dalton proposed a third law of chemical combination known as the law of multiple proportions. This law states that "when two elements A and B form more than one compound, the amounts of A that are combined with a fixed amount of B are in a small whole-number ratio." For example, carbon and oxygen form two compounds: carbon monoxide (CO) and carbon dioxide (CO2), so the masses of oxygen that combined with fixed amount of carbon stand in the ratio of 2:1. Atomic Weights A very important aspect of Dalton's work is his attempt to determine the relative masses of atoms. Water is a compound which consists of 88.8% oxygen and 11.2% hydrogen by mass. It was fatherly found that one atom of oxygen is combined with two atoms of hydrogen. The oxygen atom therefore has a mass that is approximately 8 times the mass of two 2 hydrogen atoms or 16 times the mass of one hydrogen atom. Thus the oxygen atom has a relative mass of approximately 16. It must be noted that two types of information are needed to apply Dalton's method successfully: (i) the combining ratio by mass of the elements in a compound and (ii) the combining ratio by numbers of atoms of these elements. The atomic combining ratios were not available to the chemists of Dalton's time and many years passed before the methods of chemical analysis were discovered and accurate values of atomic masses were determined. At the present time they are determined by the use of instruments known as mass spectrophotometers rather than by means of chemical analysis. Any relative atomic weight scale must be based on the arbitrary assignment of a value to one atom that is chosen as a standard. Dalton used the hydrogen atom as standard and assigned its value of 1. In latter years, chemists used naturally occurring oxygen as the standard and set its atomic weight to exactly 16. Today, a particular type of carbon atom, carbon-12 atom is employed as the standard and assigned a mass of exactly 12. An element may occur in nature as a mixture of various types of atoms that have identical chemical properties but differ slightly in mass (isotopes). The atomic weight of such an element is an average values that takes into account the masses of these types of atoms and their relative abundances in nature. Formulas The chemical symbols that are assigned to the elements are used to write formulas that describe the atomic composition of compounds. The formula 3 of water is H2O, which indicates that there are two atoms of hydrogen for every one atom of oxygen in the compound. The subscripts of the formula indicate the relative number of atoms of each type that are combined. If a symbol has no subscript, the number 1 is assigned. A molecule is a particle formed from two or more atoms. Many (but not all) compounds occur in molecular form. In these cases, the formula gives the number of atoms of each type in a single molecule of the compound. Formulas of this type are sometimes called molecular formula. Both water and sulfuric acid, for examples, has the molecular formula H2O and H2SO4 respectively. The molecular formula of hydrogen peroxide, H2O2, indicates that there are two atoms of oxygen and two atoms of hydrogen in each molecule. Notice that the ratio of hydrogen atoms to oxygen atoms (2:2) is not the simplest whole number ratio (which is 1:1). A formula that is written using the simplest whole number ratio is called the simplest formula or empirical formula. The molecular formula of hydrogen peroxide is H2O2 while the empirical formula is HO. For some compounds, the molecular and empirical formulas are the same; for examples: H2O, H2SO4, CO2 and NH3. However, for many molecular compounds molecular and empirical formulas are different for examples compound Molecular formula Empirical formula hydrazine N2H4 NH2 benzene C6H6 CH Ethane C2H6 CH3 dichloroethane C2H4Cl2 CH2Cl 4 Some compounds are not composed of molecules. For example sodium chloride is made up of ions (particles that bear positive or negative charges). The sodium and chloride ions form a crystal which composed of large numbers of these ions held together by the attractions between the dissimilar charges of the ions. In the crystal, there is one sodium ion (Na+) for every one chloride ion (Cl-). The formula of the compound is NaCl which does not describe a molecule nor does it indicate that ions are paired. Rather, the formula gives the simplest ratio of Na and Cl atoms that produce the compound. The formula NaCl, therefore, is an empirical formula. The formulas of ionic compounds are derived from the formulas of their ions. Since a crystal is electrically neutral, the total charge of positive ions must equal the total charge of the negative ions. For example in barium chloride, the formula of barium ion is Ba2+ and that of chloride ion is Cl-. The Ba2+ ion and the 2Cl- ions form BaCl2. The Mole The atomic weight of fluorine (F) is 19.0 and of hydrogen is 1.0, which means that an atom of fluorine is 19 times than atom of hydrogen. If we take 100 fluorine atoms and 100 hydrogen atoms, the mass of the collection of fluorine atoms will be 19 times the mass of the collection of hydrogen atoms which is the ratio of their atomic weights. Now, suppose that we take 19.0 g of fluorine and 1.0 g of hydrogen, the sample must contain the same number of atoms. In general, a sample of any element that has a mass in grams numerically equal to the atomic weight of the element will contain the same number of atoms. This number is called Avogadro's number. The value of Avogadro's number has been experimentally determined to be 6.02x1023. 5 The amount of substance that contains Avogadro's number of element is called mole. The atomic weight of beryllium (Be), for example, is 9.01, thus 9.01 g of Be is a mole of Be and contain 6.02x1023 beryllium atoms. Examples: 1-How many moles of aluminum (Al) are there in 125 g of Al? (the atomic weight of Al is 27.0) Solution: Since one mole of Al equals 27 g So X moles of Al equal 125 g Then the number of moles of Al = 2- How many moles and how many molecules are present in (a) 0.2 g of H2, (b) 36 g of H2O and (c) 98 g of H2SO4? 4- How many grams of gold (Au) constitute 0.25 mole of Au? (At.Wt. of Au = 197.0) Solution: Hence 1 mole of Au = 197.0 g So 0.25 moles of Au = X g of gold 0.25molofAux197.0 g Then the No. of grams (X) of gold = = 49.25 g Au. 1molofAu 5- How many carbon atoms are there in 1.00 carat diamond?. Diamond is pure carbon and one carat is exactly 0.2 g ( atomic weight of carbon = 12.01). Solution The question is: how many carbon atoms are there in 0.2 g of carbon? 6 Hence one mole of carbon = 12.01 g So, 12.01 g of carbon contains Avogadro's number of carbon atoms and, 0.200 g of C contains number equals X of C atoms so, the number of C atoms (X) in 0.200 g carbon = 0.200 gofCx6.022 x10 23 Catoms = 1.003x1022 C atoms. 12.01gofC What about compounds? The formula weight of a substance is the sum of the atomic weights of all the atoms in the formula of the substance. For example the formula weight of NaOH = 1x23 + 1x16 + 1x1 = 40. Also the formula weight of BaCl2.2H2O = 1x137.3 + 2x35.5 + 4x1 + 2x16 = 244.3. A molecular weight is the sum of the atomic weights of the atoms that constitute a molecule. The formula weight of H2O is 18 which is also the molecular weight of the substance since the formula describes the composition of water molecule. In case of NaCl, the formula weight is not the molecular weight since NaCl is an ionic compound and molecules of NaCl do not exist. A mole of molecular substance consists of Avogadro's number of molecules. A mole of H2O, therefore, has a mass of 18.0 g contains 6.022x1023 molecules. Since there two atoms of hydrogen and one atom of oxygen in H2O molecule, a mole of H2O molecules two moles of hydrogen atoms (2.0 g) and one mole of oxygen atom (16.0 g). Derivation of Formulas 7 Data from chemical analysis of a compound are used to derive the empirical formula of the compound. The analysis gives the proportions by mass of the elements that make up the compound. The simplest or empirical formula indicates the relative numbers of atoms of various types that make up the compound. The derivation of empirical formula can be obtained following the following sequences: 1- If the data are given in terms of percentage composition, base the calculation on a 100 g sample of the compound. In this case, the number of grams of each element is equal to the percentage of the element present in the compound. There is no need to find percentage of the element if the data are given in terms of number of grams. 2- Convert the number of grams of each element to the number of moles of atoms of each atom. The number of moles equals the number of grams divided by the atomic weight. 3- Divide each of the values obtained in step 2 by the smallest value. If every number obtained in this way is not a whole number multibly each number by the same simple integer in such a way that whole number will result. 4- The whole numbers obtained in step 3 are the subscripts of the empirical formula. Examples: 6- What is the empirical formula of compound that contains 43.6% P and 56.4% O? ( atomic weights of P and O are 31.0 and 16.0 respectively). Solution: 8 Assume that the sample has a mass of 100.0 g, thus there are 43.6 g of P and 56.4 g of O The number of moles of P = 43.6 g / 31.0 = 1.41 mol P The number of moles of O = 56.4 g / 16.0 = 3.53 mol O Element P O No. of moles 1.41 3.53 dividing by smallest value 1 2.5 multibly by 2 to get whole number 2 5 Thus the empirical formula is P2O5. 7- Caffeine, which occurs in coffee, tea and kola nuts, is a stimulant for the central nervous system. A 1.261 g sample of pure caffeine contains 0.624 g C, 0.065 g H, 0.364 g N and 0.208 g O. what is the empirical formula of caffeine? Solutions: Elements C H N O No. of moles 0.052 0.065 0.026 0.013 dividing by smallest value 4 5 2 1 (Whole numbers are obtained) Thus the empirical formula is C4H5N2O. The molecular formula of a compound can be derived from the empirical formula if the molecular weight of the compound is known The molecular weight = n x empirical formula weight, where n is a whole number. Example 8: One phosphorous oxide has an empirical formula P2O5, its molecular weight is 284. What is the molecular formula? Solution: 9 The empirical formula weight = 2 x 31 + 5 x 16 = 142 Thus n = 284 / 142 = 2 Therefore, there are twice as many atoms of each kind present in the molecule, so the molecular formula is P4O10. Example 9: The molecular weight of caffeine is 194 and the empirical formula is C4H5N2O. What is the molecular formula? Example 10: The compound glucose contains 40.0% C, 6.73% H and the rest is O. If its molecular weight is 180.2, what is the molecular formula? Percentage Composition of Compounds The percentage composition of a compound is readily calculated from its formula and from the atomic weights of the elements constitute the compound. The process is illustrated by the following examples. Example 11: What is the percentage of Fe (At.Wt. 55.8) in Fe2O3? Solution One g mole of Fe2O3 = 2x55.8 + 3x16 = 159.6 g 2 x55.8 The % of Fe = x100  69.92% Fe in Fe2O3. 159.6 The % of O can be calculated in a similar way or by difference from 100%. If the formula is not known, the percentage composition is frequently determined by chemical analysis. These data can be used to find the empirical formula of the compound. Example 12: Nicotine is a compound that contains carbon, hydrogen and nitrogen. If 2.50 g sample of nicotine is burned in oxygen, 6.78 g of CO2 , 11 1.94 g of H2O and 0.432 g of N2 were produced. What is the percentage composition of nicotine? Solution: We first calculate the quantity of each element present in the 2.5 g sample. The C forms 6.78 g of CO2, thus One C forms one mole of CO2 Then 12 g C forms 44 g of CO2 6.78 x12 X g C forms 6.78 g of CO2 So, X = 1.85 g C 44 1.85 gC The % of C = x100  74.0% C in nicotine 2.50 gni cot ine In a similar way: 18 g of H2O form 2.02 g of H, then 1.94 x2.02 1.94 g H2O form X g of H So, X =  0.218 g H 18.0 0.218 gH The % of H = x100  8.72 % H in nicotine 2.50 gni cot ine 0.432 gN Finally, the % of N = x100  17.3% % N in nicotine. 2.5 gNi cot ine Example 13: Find the empirical formula of nicotine. Example 14: Silver sulfide (Ag2S) occurs as the mineral argentite, which is an ore of silver. How many grams of silver are theoretically obtained from 250.0 g of an impure ore that is 70.00 % Ag2S? Example 15: What mass of copper (Cu) is theoretically obtained from 10.0 kg of chalcocite ore that is 25.0 % Cu2S? Chemical Equations 11 Chemical equations are representation of reactions in terms of symbols and formulas. They report the results of experimentation, for example carbon disulfide (CS2) reacts with chlorine (Cl2) to produce carbon tetra- chloride (CCl4) and disulfur dichloride (S2Cl2). This is represented as: CS2 + Cl2 → CCl4 + S2Cl2 This equation is not quantitatively correct because only two chlorine atoms enter the reaction and six chlorine atoms produced which violates the law of conservation of mass. To balance the equation the number of chlorine molecules must be multiblied by three, so CS2 + 3Cl2 → CCl4 + S2Cl2 The simplest types of chemical equations are balanced by trial and error. Example 16: Represent the following reactions by balanced chemical equations; 1- Reaction of steam with hot iron (Fe) to give magnetite (Fe3O4) and H2 2- Complete combustion of ethane (C2H6) to give CO2 and water vapor. Problems based on chemical equations Chemical equation can be interpreted in several different ways, for example the equation: 2H2 + O2 → 2H2O shows that two moles of H react with one mole of O to give two moles of water. In other way, it also shows that 2.02 g of H react with 32.0 g of O to give 36.0 g of water. Example 17: Determine the number of moles of O2 that reacted with 5.0 moles of ethane according to the equation: 2C2H6 + 7O2 → 4CO2 + 6H2O Solution: It is clear from the balanced equation that 12  7 moles of O2 is required to react with 2.0 moles of ethane so, X moles of O2 is required to react with 5.0 moles of ethane 5 x7 thus, the No. of moles of O2 (X) =  17.5 mol O2. 2 Example 18: Chlorine can be prepared by the reaction: MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O (a): how many grams of HCl are required to react with 25.0 g of MnO2? (b): how many grams of Cl2 are produced by the reaction? Example 19: The amount of carbon monoxide (CO) in a sample of gas is determined by the reaction: I2O5 + 5CO → I2 + 5CO2. If a gas sample liberates 0.192 g of I2, how many grams of CO were present in the sample? The limiting reactant In some cases, quantities are given for two or more reactants. For example when asking: how many moles of H2O can be prepared from 2 moles of H and 2 moles of O2 by the equation: 2H2 + O2 → 2H2O. In such case more O2 has been supplied than can be used. When all H2 has been consumed, the reaction will stop. Thus the amount of H2 supplied limits the reaction and is called the limiting reactant. Whenever the quantities of two or more reactants are given, we must determine which one limits the reaction. This can be done as follows: 1- Calculate the number of moles of each of the given reactants. 2- Divide each value by the coefficient of the chemical equation. 3- the smallest number obtained pertains to the limiting reactant. Use the quantity of this reactant to solve the problem. 13 Example 20: How many moles of H2 can be theoretically prepared from 4.00 mol of Fe and 5.00 mol of H2O according to the equation: 3Fe + 4H2O → Fe3O4 + 4H2 Solution: First determine the limiting reactant 1- For Fe: the number of moles given in the problem is 4.00 and the coefficient of Fe in chemical equation is 3. Thus the amount of Fe given is 4.00 mol Fe / 3mol Fe = 1.33 times the amount specified in the chemical equation. 2- In a similar case, the amount of H2O is 5.00 mol H2O / 4 mol H2O = 1.25 times the amount specified in the chemical equation. From I & 2, the limiting reactant is the H2O. 4molH 2 Thus, the No. of moles of H2 = 5.00 mol H2Ox  5.0 mol H2. 4molH 2O Example 21: How many grams of N2F4 can be theoretically prepared from 4.00 g of NH3 and 14.0 g of F2?. The chemical equation for the reaction is: 2NH3 + 5F2 → N2F4 + 6HF Percent yield: Actually the quantity of the product obtained from the reaction is less than the amount calculated. This is due to part of the reactants do not react, or this part of the reactants react in a different way (side reactions) or that not all of the product is recovered. The percent yield is given by: actualyield Percent yield = x100 % theoretica l 14 Stoichiometry of Reactions in Solution Many chemical reactions are carried out in aqueous solutions. The quantities of reactants in this case are usually stated in terms of solution concentration. The molarity (M) of a solution is the number of moles of substance (solute) dissolved in one liter of solution. The relation between mass (m) of dissolved solute, molarity (M) and the volume (V) in ml of solution is given by: MxMolecularweightxvolume Mass (g) = 1000 While the number of moles (n) is given by mass n= mol molarmass Example 22: How many moles of AgNO3 are present in 25.0 ml of 0.600 M AgNO3 solution? Example 23: How many grams of NaOH are required to prepare 0.250 liter of 0.300 M solution? Example 24: Soda mint tablet contains NaHCO3 as an antacid. One tablet requires 34.5 ml of 0.138 M HCl solution for complete reaction. Determine the number of grams of NaHCO3 that one tablet contains. 15