The Mole Concept Unit-1 (Honors Chem 2024) PDF

THE MOLE UNIT (COMPOUND Honors Chem 2024 CALCULATIONS) AVERAGE ATOMIC Part 1 MASS  ATOMIC MASS We have talked about the idea of isotopes. When we take a sample of carbon, how do we know if we have C-12, C-13, or C- 14? We don’t, so we work with the average atomic mass, whic...

THE MOLE UNIT (COMPOUND Honors Chem 2024 CALCULATIONS) AVERAGE ATOMIC Part 1 MASS  ATOMIC MASS We have talked about the idea of isotopes. When we take a sample of carbon, how do we know if we have C-12, C-13, or C- 14? We don’t, so we work with the average atomic mass, which takes into account how ABOUT THE UNITS Since atoms are so tiny, 1 amu = 1.66x10-24 g scientists came up with their own unit of atomic mass, the This is roughly equal to the atomic mass unit (amu - mass of a proton (1.673x10-24 g) although some people or a neutron (1.675x10-24 g) abbreviate it to u). They decided to make the amu based on a carbon-12 atom (the basis of life as we know it. Personally, I think they should have picked hydrogen-1, but I wasn’t invited to that party. THE EQUATION For example, there are two stable isotopes of boron. B-10 has a mass of 10.01 amu and makes up 19.9% of all stable boron atoms. B-11 has a mass of 11.01 amu and makes up 80.1% of all stable boron atoms. Let’s find its average atomic mass: PRACTICE PACKET #1 An unknown element has two stable isotopes. The first has a mass of 14.003074 amu and makes up 99.63% of all atoms of this element. The second has a mass of 15.000108 amu. What is the average atomic mass of this element? What is the element?  ANSWER The element must be nitrogen  PRACTICE PACKET #2 Atomic A new Isotope Mass % theoretical Abundance (amu) element is J-270 270.600 36.45 determined to J-271 271.700 20.77 have 3 stable isotopes. Atomic J-272 272.800 42.78 masses and abundances are given below. What is the ANSWER Copper is made up of two isotopes, Cu-63 (62.9296 amu) and Cu-65 (64.9278 amu). Given that copper's average atomic mass is 63.546 amu, what is the percent abundance of each isotope? To get credit you must specify which isotope has which percent abundance. HOW ABOUT A BONUS OPPORTUNITY? ANSWER  PRACTICE PACKET #4 A sample of a naturally occurring element consists of two isotopes. X-240 (mass = 240.240 amu) and X-243 (mass = 243.534 amu). If the average atomic mass of this element is 241.800 amu, what is the percent abundance of X- 243? ANSWER THE MOLE Part 2  A THOUGHT QUESTION If an atom of hydrogen is 1.0 amu, how many hydrogen atoms would it take to have exactly 1.0 g of hydrogen atoms? THE ANSWER IS: 6.02 X 10 23 ATOMS We said that 1 amu = 1.66 x 10-24 g This number is referred to as a mole, it is also called Avogadro’s #, named after the Italian chemist Amadeo Avogadro who came up with the idea of the number. SO, WHAT EXACTLY IS A MOLE? What is a dozen? We know that a dozen means 12 of whatever we are talking about. A dozen eggs is 12 eggs, a dozen cars is 12 cars, and a dozen people is 12 people. Likewise, a mole means 6.02x10 23 of whatever we are talking about. A mole of carbon is 6.02x10 23 carbon atoms, a mole of water is 6.02x1023 water molecules, and a mole of sand is 6.02x1023 grains of sand.  PRACTICE PACKET #5 An amu is an artificial unit used to describe the mass of atoms. It is based on a C-12 atom, under the assumption that a C-12 atom must have a mass of exactly 12.00 amu. By determining the mass of a C-12 atom we have thus calculated that an amu = 1.66 x 10- 24 g. a) What is the mass of a single C-12 atom in grams? ________________________________________ b) If I have a sample of pure C-12 which has a mass of exactly 12.00 g, how many atoms of C-12 must be in the sample? ________________________________ (this  ANSWERS An amu is an artificial unit used to describe the mass of atoms. It is based on a C-12 atom, under the assumption that a C-12 atom must have a mass of exactly 12.00 amu. By determining the mass of a C-12 atom we have thus calculated that an amu = 1.66 x 10- 24 g. a) What is the mass of a single C-12 atom in grams? b) If I have a sample of pure C-12 which has a mass of exactly 12.00 g, how many atoms of C-12 must be in the sample? MORE ABOUT THE MOLE Atoms are tiny, tiny particles, too small to be counted. The idea of a mole is a means of counting atoms based on weighing them; at the arcade at the boardwalk, they used to count the tokens you won by finding their mass; knowing the mass of one token allows the computer to calculate how many tokens there are without actually counting them. Think about the magnitude of the mole: 6.02x1023 atoms = 602,000,000,000,000,000,000,000 If you lived to be 1000 years old, you would have lived for 31,557,600,000 seconds, that is a tiny, tiny fraction of a mole of seconds. PRACTICE PACKET #6 If I have 3.00 moles of Ag atoms, how many Ag atoms do I have?  ANSWER: 𝟐𝟑 𝟔.𝟎𝟐𝒙𝟏𝟎 𝒂𝒕𝒐𝒎𝒔𝑨𝒈 𝟐𝟒 𝟑.𝟎𝟎𝒎𝒐𝒍𝑨𝒈 =𝟏.𝟖𝟏𝒙𝟏𝟎 𝒂𝒕𝒐𝒎𝒔𝑨𝒈 𝟏𝒎𝒐𝒍𝑨𝒈 PRACTICE If I have 10.0 moles of PACKET Mo atoms, how many #7 Mo atoms do I have?  ANSWER: 𝟐𝟑 𝟔.𝟎𝟐𝒙𝟏𝟎 𝒂𝒕𝒐𝒎𝒔𝑴𝒐 𝟐𝟒 𝟏𝟎.𝟎𝒎𝒐𝒍𝑴𝒐 =𝟔.𝟎𝟐𝒙𝟏𝟎 𝒂𝒕𝒐𝒎𝒔𝑴𝒐 𝟏𝒎𝒐𝒍𝑴𝒐 PRACTICE PACKET #8 If I have 4.81 x 1024 atoms of Ba, how moles of Ba atoms do I have?  ANSWER: 𝟐𝟒 𝟏𝒎𝒐𝒍𝑩𝒂 𝟒.𝟖𝟏𝒙𝟏𝟎 𝒂𝒕𝒐𝒎𝒔𝑩𝒂 𝟐𝟑 =𝟕.𝟗𝟗𝒎𝒐𝒍𝑩𝒂 𝟔.𝟎𝟐𝒙𝟏𝟎 𝒂𝒕𝒐𝒎𝒔𝑩𝒂 If I have 3.75 x 1024 atoms of B, how moles of B atoms do I have? PRACTICE PACKET #9  ANSWER: 𝟐𝟒 𝟏𝒎𝒐𝒍𝑩 𝟑.𝟕𝟓𝒙𝟏𝟎 𝒂𝒕𝒐𝒎𝒔𝑩 𝟐𝟑 =𝟔.𝟐𝟑𝒎𝒐𝒍𝑩 𝟔.𝟎𝟐𝒙𝟏𝟎 𝒂𝒕𝒐𝒎𝒔𝑩 TWO ASPECTS OF THE MOLE A mole is a number of A mole also represent a atoms, molecules, ions, specific mass of an element. electrons, etc. 1 mol of C = 12.011 g 1 mol of C = 6.02x1023 atoms of C 1 mol of O = 15.9994 g 5 mol of C = 3.01 x 1024 1 mol of Mo = 92.94 g atoms of C We talked about the atomic 0.8 mol of C = 4.82 x 1023 mass, which is the mass of atoms of C one atom. Now we are calling the same number something else: the molar mass – the mass of one mole of a substance. PRACTICE PACKET #10 One atom of carbon has an average mass of _________________ amu. One mole of carbon atoms has a mass of ____________________ g.  ANSWERS One atom of carbon has an average mass of 12.011 amu. One mole of carbon atoms has a mass of 12.011 g.  PRACTICE PACKET #10 One atom of tungsten has an average mass of _________________ amu. One mole of tungsten atoms has a mass of _________________ g. 5.00 moles of tungsten would have a mass of _________________ g. 0.250  ANSWERS One atom of tungsten has an average mass of 183.84 amu. One mole of tungsten atoms has a mass of 183.84 g. 5.00 moles of tungsten would have a mass of 919 g. 0.250 moles of tungsten would have a mass of 46.0 g. PRACTICE PACKET #12 Determine the molar mass of the following elements or compounds: silver ____________________________ molybdenum ____________________________ boron ____________________________ barium ____________________________ permanganic acid ____________________________ potassium oxalate ___________________________ tetranitrogen decoxide ______________________ octene ____________________________  ANSWERS Determine the molar mass of the following elements or compounds: silver 107.9 g/mol molybdenum 92.94 g/mol boron 10.8 g/mol barium 137.3 g/mol permanganic acid 119.9 g/mol potassium oxalate 166.2 g/mol tetranitrogen decoxide 216.0 g/mol octene 112.2 g/mol  PRACTICE PACKET #13 One molecule of octene has a mass of _______ amu. One mole of octene has a mass of _______ g. 2.00 moles of octene would have a mass of _____ g. 0.0125 moles of octene would have a  ANSWERS One molecule of octene has a mass of 112.2 amu. One mole of octene has a mass of 112.2 g. 2.00 moles of octene would have a mass of 224 g. 0.0125 moles of octene would have a mass of 1.40 g. PRACTICE PACKET #14 What is the mass, in grams, of: a) 4.00 moles of calcium? b) 1.24 moles of lead? c) 0.00750 moles of potassium oxalate? d) 22.2 moles of tetranitrogen decoxide?  ANSWERS What is the mass, in grams, of: a) 4.00 moles of calcium? b) 1.24 moles of lead? c) 0.00750 moles of potassium oxalate? d) 22.2 moles of tetranitrogen decoxide? PRACTICE PACKET #15 One mole of gold has a mass of _________ g. If I have 39.4 grams of gold, how many moles of gold atoms do I have?  ANSWER One mole of gold has a mass of 197.0 g. If I have 39.4 grams of gold, how many moles of gold atoms do I have? PRACTICE PACKET #16 One mole of sodium has a mass of __________ g. If I have 321 grams of sodium, how many moles of sodium atoms do I have?  ANSWER One mole of sodium has a mass of 23.0 g. If I have 321 grams of sodium, how many moles of sodium atoms do I have? PRACTICE PACKET #17 One mole of permanganic acid has a mass of _________ g. If I have 1199 grams of permanganic acid, how many moles do I have?  ANSWER One mole of permanganic acid has a mass of 119.9 g. If I have 1199 grams of permanganic acid, how many moles do I have?  PRACTICE PACKET #18 How many moles are in a sample of: a) 16.0 grams of cobalt? b) 111 grams of zinc? c) 2.84 grams of platinum? d) 311 g of octene? e) 5.33 g of potassium oxalate?  PRACTICE PACKET #18 How many moles are in a sample of: a) 16.0 grams of cobalt? b) 111 grams of zinc? c) 2.84 grams of platinum? d) 311 g of octene? e) 5.33 g of potassium oxalate?  TWO STEP PROBLEMS If we want to go from grams to atoms, or atoms to grams, we have to go to moles first, so even though moles are not mentioned, it is the heart of everything in this unit. For example, how many atoms are in 4.80 g of C?  PRACTICE PACKET #19 How many atoms are in a sample of: a) 16.0 grams of cobalt? b) 111 grams of zinc? c) 2.84 grams of platinum?  ANSWERS How many atoms are in a sample of: a) 16.0 grams of cobalt? b) 111 grams of zinc? c) 2.84 grams of platinum? How many molecules are in a sample of 82.4 g of tetranitrogen decoxide? PRACTICE PACKET #20  ANSWER How many molecules are in a sample of 82.4 g of tetranitrogen decoxide?  PRACTICE PACKET #21 How many atoms are in a sample of 82.4 g of tetranitrogen decoxide?  ANSWER How many atoms are in a sample of 82.4 g of tetranitrogen decoxide? How many grams are there in a sample of 4.19 x 10 molecules 22 of permanganic acid? PRACTICE PACKET #22  ANSWER How many grams are there in a sample of 4.19 x 1022 molecules of permanganic acid? PRACTICE PACKET #23 How many atoms are in a sample of 20.0 grams of calcium?  ANSWER How many atoms are in a sample of 20.0 grams of calcium? WHAT IS THE MASS, IN GRAMS, OF 3.83 X 10 25 PRACTICE PACKET #24 ATOMS OF LITHIUM?  ANSWER What is the mass, in grams, of 3.83 x 1025 atoms of lithium? PRACTICE PACKET #25 What is the mass, in grams, of 8.173 x 10 molecules of 22 potassium oxalate?  ANSWER What is the mass, in grams, of 8.173 x 1022 molecules of potassium oxalate?  PRACTICE PACKET #26 A spherical ball of iron, has a density of 7.874 g/ml and a radius of 47.0 mm. How many atoms of iron are present?  I LOVE THIS QUESTION! A spherical ball of iron, has a density of 7.874 g/ml and a radius of 47.0 mm. How many atoms of iron are present? Remember that ml = cm3, so we want to convert the radius to cm.  PERCENT COMPOSITION Part 3  PERCENT COMPOSITION Also called mass percent, or percent by mass, the percent composition describes the composition of a compound based on the amount of mass each element contributes. It all starts with the molar mass. For example, let’s consider H2O 2H= 2(1.0 g/mol) = 2.0 g/mol 1 O = 1(16.0 g/mol) = 16.0 g/mol The total molar mass is 18.0 g/mol Hydrogen contributed 2.0 g out of every 18.0 g Oxygen contributed 16.0 g out of every 18.0 g  PRACTICE PACKET #27 Find the percent composition of each of the following compounds: A) carbonic acid B) carbon dioxide C) butane D) dichromic acid E) arsenic triiodide F) antimony (V) phosphite LET’S DO TWO AT A TIME a) Carbonic acid – H2CO3 Carbon dioxide – CO2 2 H = 2(1.0 g/mol) = 2.0 1 C = 1(12.0 g/mol) = 12.0 g/mol g/mol 1 C = 1(12.0 g/mol) = 12.0 2 O = 2(16.0 g/mol) = 32.0 g/mol g/mol 3 O = 3(16.0 g/mol) = 48.0 g/mol O O THE NEXT TWO Butane – C4H10 Dichromic acid – H2Cr2O7 4 C = 4(12.0 g/mol) = 48.0 2 H = 2(1.0 g/mol) = 2.0 g/mol g/mol 10 H = 10(1.0 g/mol) = 10.0 2 Cr = 2(52.0 g/mol) = 104.0 g/mol g/mol 7 O = 7(16.0 g/mol) = 112.0 g/mol H O THE LAST TWO Antimony (V) phosphite – Arsenic triiodide – AsI3 Sb3(PO3)5 1 As = 1(74.9 g/mol) = 74.9 3 Sb = 3(121.8 g/mol) = g/mol 365.4 g/mol 3 I = 3(126.9 g/mol) = 380.7 5 P = 5(31.0 g/mol) = 155.0 g/mol g/mol 15 O = 15(16.0 g/mol) = 240.0 g/mol As: Sb: I: P: O: EMPIRICAL FORMULA Part 4  For our purposes, empirical means simplified, as in whole number ratio. For example, we know that the formula of butane is C4H10. That is called the molecular formula (the actual formula). If we simplify the ratio (dividing each element by two in this Why case), would wewe get an want theempirical empirical DEFINITI formula when of C2Hwe 5. know the actual (molecular) formula? We wouldn’t. ON However, experimentally, it is easier to determine the empirical formula first and then use that as a means to finding the molecular formula. In other words, this is the first step to determining the molecular formula.  1) Assume that we have 100 grams of the STEPS TO compound; in other words, change the percents to grams. FINDING THE EMPIRICA 2) convert the masses to moles. L FORMULA 3) simplify the ratio of moles by dividing each number of moles by the smallest FROM THE value. PERCENT 4) If simplified moles are not reasonably close to whole numbers (within a few COMPOSI hundredths), multiply each value by a number (typically 2-4) that will yield all TION whole numbers.  A compound LET’S TAKE is 30.44 % nitrogen THIS STEP BY and 69.56% STEP THROUGH oxygen. AN EXAMPLE Determine the empirical formula. 1) ASSUME THAT WE HAVE 100 GRAMS OF THE COMPOUND; IN OTHER WORDS, CHANGE THE PERCENTS TO GRAMS. 30.44 % N = 30.44 g N 69.56 % O = 69.56 g O 2) CONVER T THE MASSES TO MOLES.  3) SIMPLIFY THE RATIO OF MOLES BY DIVIDING EACH NUMBER OF MOLES BY THE SMALLEST VALUE. Since these are both reasonably close to whole numbers, we have found our A compound is 84.12 % carbon and 15.88% hydrogen. Determine the empirical formula. LET’S DO ONE MORE EXAMPLE 1) ASSUME THAT WE HAVE 100 GRAMS OF THE COMPOUND; IN OTHER WORDS, CHANGE THE PERCENTS TO GRAMS. 84.12 % 15.88 % C= H= 84.12 g 15.88 g C H 2) CONVERT THE MASSES TO MOLES.  3) SIMPLIFY THE RATIO OF MOLES BY DIVIDING EACH NUMBER OF MOLES BY THE SMALLEST VALUE. 15.88 Since these are not both reasonably close to whole numbers, we need to multiply both values by something to make them whole numbers. Maybe 4?  4) IF SIMPLIFIED MOLES ARE NOT REASONABLY CLOSE TO WHOLE NUMBERS (WITHIN A FEW HUNDREDTHS), MULTIPLY EACH VALUE BY A NUMBER (TYPICALLY 2-4) THAT WILL YIELD ALL WHOLE NUMBERS. The empirical formula is C4H9 PRACTICE PACKET #28 A compound is 80.0% carbon and 20.0% hydrogen. What is its empirical formula? ANSWER: CH 3 PRACTICE PACKET #29 A compound is 27.29% carbon and 72.71% oxygen. What is its empirical formula? ANSWER: CO2 PRACTICE PACKET #30 A compound is 25.94% nitrogen and 74.06% oxygen. What is its empirical formula? ANSWER: N2O5  A compound is 3.25% hydrogen, PRACTICE 19.36% carbon and PACKET #31 77.39% oxygen. What is its empirical formula? ANSWER: H2CO3  PRACTICE PACKET #32 A compound is found to be composed of 20.19% magnesium, 26.64% sulfur, and 53.17% oxygen. What is the empirical formula of this compound? ANSWER: MgSO4 MOLECULAR FORMULA Part 5  As previously stated, the molecular formula is the actual formula. When we ABOUT learned nomenclature, we THE were always talking about the molecular formula. MOLECUL Let’s look at the difference AR between the two types of FORMULA formulas before we figure out how to find the molecular formula. Write the empirical formula for the following compounds. a) C6H6 b) C8H18 PRACTICE PACKET c) WO2 #33 d) C2H6O2 e) X39Y13  ANSWERS a) C6H6 CH different by a factor of 6 b) C8H18 C4H9 different by a factor of 2 c) WO2 WO2 not different (factor of 1) d) C2H6O2 CH3Odifferent by a factor of 2 e) X39Y13 X3Y different by a factor of 13 The trick to finding the molecular formula, from the empirical formula, is to determine what that factor is, differentiating the two DETERMI formulas. NING THE MOLECUL To determine the factor, and thus find AR the molecular formula, FORMULA we need to know two things: 1) the empirical formula and 2) the actual molar mass. THREE STEPS TO FINDING THE MOLECULAR FORMULA 1) Determine the molar mass of the empirical formula (I will refer to this as the empirical molar mass) 2) Determine the factor that the molecular formula is off by: 3) Multiply the empirical formula by the factor Before we found the empirical formula of a compound of nitrogen and oxygen to be N2O5. If the actual molar mass is 216.0 g/mol, what is the molecular formula? LET’S TRY ONE LET’S GO THROUGH THE PROCESS 1) Determine the molar mass of the empirical formula (I will refer to this as the empirical molar mass) N2O5 = 2 N + 5 O = 2 14.0+5 16.0 = 108.0 g/mol 2) Determine the factor that the molecular formula is off by: 3) Multiply the empirical formula by the factor 2(N2O5)= N4O10 Now that we know the molecular formula, we can name it too – tetranitrogen decoxide PRACTICE PACKET #34 A compound with an empirical formula of C2H4O has a molar mass of 132.0 g/mol. What is the molecular formula of this compound? ANSWER: C6H12O3 1) Determine the molar mass of the empirical formula (I will refer to this as the empirical molar mass) C2OH4 = 2 C + 1 O + 4 H = 2 12.0+1 16.0 + 4 1.0 = 44.0 g/mol 2) Determine the factor that the molecular formula is off by: 3) Multiply the empirical formula by the factor 2(C2OH4)= C6H12O3 PRACTICE PACKET #35 A compound with an empirical formula of C4H4O has a molar mass of 272.0 g/mol. What is the molecular formula  ANSWER: C16H16O4 C4H4O = 4 C + 4 H + 1 O = 4 12.0 + 4 1.0 + 1 16.0 = 68.0 g/mol 4(C4H4O) = C16H16O4 PRACTICE PACKET #36 A compound with an empirical formula of CBrOF has a molar mass of 254.7 g/mol. What is the molecular formula of this compound?  ANSWER: C2Br2O2F2 CBrOF = C + Br + O + F = 12.0 + 79.9 + 16.0 + 19.0 = 126.9 g/mol 2(CBrOF) = C2Br2O2F2  PRACTICE PACKET #37 A well-known reagent in analytical chemistry, dimethylglyoxime, has the empirical formula C2H4NO. If its molar mass is 116.1 g/mol, what is the  ANSWER: C4H8N2O2 C2H4NO = 2 C + 4 H + N + O = 2 12.0 + 4 1.0 + 14.0 + 16.0 = 58.0 g/mol 2(C2H4NO) = C4H8N2O2  PRACTICE PACKET #38 Nitrogen and oxygen form an extensive series of oxides with the general formula NxOy. One of them is a blue solid that comes apart, reversibly, in the gas phase. It contains 36.84% N. a) What is the empirical formula of this oxide? b) If its molar mass is roughly 228 g/mol, then what is the molecular formula of this oxide PART A) EMPIRICAL FORMULA: N 2O3  PART B) MOLECULAR FORMULA: N 6O9 N2O3 = 2 N + 3 O = 2 14.0 + 3 16.0 = 76.0 g/mol 3(N2O3) = N6O9  PRACTICE PACKET #39 An unknown compound was found to have a percent composition as follows: 47.0 % potassium, 14.5 % carbon, and 38.5 % oxygen. a) What is its empirical formula? b) If the true molar mass of the compound is 166.22 g/mol, what is PART A) EMPIRICAL FORMULA: KCO 2  PART B) MOLECULAR FORMULA: K2C2O4 KCO2 = 1 K + 1 C + 2 O = 1 39.1 + 1 12.0 + 2 16.0 = 83.1 g/mol 2(KCO2) = K2C2O4 HYDRATE ANALYSIS Part 6  WHAT IS A HYDRATE? Hydrates are ionic compounds that lock water molecules inside of their crystalline structure. If you’ve bought new shoes or electronics, there is often a packet of little crystals (usually some type of silica/silicate) inside that says don’t eat! These are hydrates, that were initially removed of their water, so that they would be “thirsty” for more, absorbing water into themselves to keep the dampness out of your purchase! Sometimes, when the water is removed from these compounds, they are referred to as anhydrous. Although they can be written in multiple ways, the most common way to write the formula of a hydrate is to write the ionic DID SOMEONE SAY # of H2 O prefix NOMENCLATURE? 1 2 mono- di- There is a little different nomenclature for hydrates, but 3 tri- it really isn’t too new. When we name a hydrate, we 4 tetra- name the ionic compound the same as always but then we add a prefix to the suffix –hydrate to indicate the 5 penta- number of water molecules. 6 hexa- On the previous slide, we looked at MgSO4 7H2O. 7 hepta- MgSO4 is, of course, called magnesium sulfate 8 octa- 7H2O is called heptahydrate 9 nona- So, MgSO4 7H2O is called magnesium sulfate 10 deca- heptahydrate  PRACTICE PACKET #40 Name the following hydrates: a. Na2S2O3 5 H2O b. CaSO4 2 H2O c. MgSO4 H2O d. Mn(NO3)2 4 H2O  ANSWERS Name the following hydrates: a. Na2S2O3 5 H2O sodium thiosulfate pentahydrate b. CaSO4 2 H2O calcium sulfate dihydrate c. MgSO4 H2O magnesium sulfate monohydrate d. Mn(NO3)2 4 H2O manganese (II) nitrate tetrahydrate  PRACTICE PACKET #41 Write the formulas of the following hydrates: a. magnesium nitrate hexahydrate b. iron (II) sulfate heptahydrate c. copper (II) nitrate trihydrate d. tin (II) chloride dihydrate  ANSWERS Write the formulas of the following hydrates: a. magnesium nitrate hexahydrate Mg(NO3)2 6H2O b. iron (II) sulfate heptahydrate FeSO4 7H2O c. copper (II) nitrate trihydrate Cu(NO3)2 3H2O d. tin (II) chloride dihydrate SnCl2 2H2O A task, very similar to finding the empirical formulas that we have done in this unit, is to determine the empirical ratio between the water and the ionic compound in the hydrate. In other words, we want to be able to figure out the number of water molecules for every unit of ionic compound. NOW FOR THE ANALYSIS PART THESE STEPS SHOULD SEEM FAMILIAR 1) Determine the mass of each compound (ionic compound and water) 2) convert the masses to moles. 3) simplify the ratio of moles by dividing each number of moles by the smallest value. EXAMPLE 1 For example, let’s consider a hydrate of calcium carbonate, CaCO3 X H2O. If this hydrate is determined to contain 3.571 g CaCO3 and 6.429 g H2O, determine the exact formula of this hydrate.  ANSWER: CaCO3 10H2O calcium carbonate decahydrate 1) Determine the mass of each compound. This was done for us this time. 2) Convert the mass to moles. 3) Simplify the ratio by dividing by the smaller number of moles  WHAT IS THE FORMULA AND THE NAME FOR A HYDRATE PRACTIC THAT IS 90.7g E PACKET SrC2O4 AND #42 9.30g H2O?  ANSWER: H2O strontium oxalate monohydrate 1) Determine the mass of each compound. This was done for us this time. 2) Convert the mass to moles. 3) Simplify the ratio by dividing by the smaller number of moles  What is the formula and the name of a hydrate PRACTICE that is PACKET #43 86.7% Mo2S5 and 13.3% H2O?  ANSWER: molybdenum (V) sulfide trihydrate 1) Determine the mass of each compound. This was done for us this time. Assume we have 100.0 g of the hydrate 86.7 % of 100.0 g = 86.7 g Mo2S5 13.3% of 100.0 g = 13.3 g H2O 2) Convert the mass to moles. 3) Simplify the ratio by dividing by the smaller number of moles EXAMPLE 2 In a lab, 10.000 grams of a hydrate Fe(ClO3)3 XH2O was heated, afterwards only 8.096 g of the anhydrate remain. What is the name and the formula of this hydrate?  ANSWER: iron (III) chlorate tetrahydrate 1) Determine the mass of each compound. This was done for us this time. We know that we have 8.096 g of the Fe(ClO3)3 10.0 g Fe(ClO3)3 XH2O – 8.096 g Fe(ClO3)3 = 1.904 g H2O 2) Convert the mass to moles. 3) Simplify the ratio by dividing by the smaller number of moles During lab, 1.62 g of CoCl2 XH2O were heated. After heating, only 0.88 g of CoCl2 remained. What was the formula and the name of the original hydrate? PRACTICE PACKET #44  ANSWER: cobalt (II) chloride hexahydrate 1) Determine the mass of each compound. This was done for us this time. We know that we have 0.88 g of the CoCl2 1.62 g CoCl2 XH2O – 0.88 g CoCl2 = 0.74 g H2O 2) Convert the mass to moles. 3) Simplify the ratio by dividing by the smaller number of moles PRACTICE PACKET #45 During lab, 1.04 g of NiSO4 __ H2O were heated. After heating, only 0.61 g of NiSO4 remained. What was the formula and the name of the original hydrate?  ANSWER: nickel (II) sulfate hexahydrate 1) Determine the mass of each compound. This was done for us this time. We know that we have 0.61 g of the NiSO4 1.04 g NiSO4 XH2O – 0.61 g NiSO4 = 0.43 g H2O 2) Convert the mass to moles. 3) Simplify the ratio by dividing by the smaller number of moles  MASS PERCENT OF WATER IN A HYDRATE This is exactly like percent composition, except we are considering water as one piece (not separate elements). For example, if we consider the hydrate of the last problem, NiSO4 6H2O: Ni = 58.7 = 58.7 S = 32.1 = 32.1 4 O = 4 16.0 = 64.0 6 H2O = 6 18.0 = 108.0 262.8 g/mol  When we solved #45, we had found that 1.04 g of the hydrate was 0.61 g of NiSO4 and 0.43 g FINDING water. This means that the water was THE MASS 0.43 g/1.04 g or 41.3% of the % hydrate. The only reason why we got 41.1 ANOTHER % using the molar mass and WAY 41.3% here is that the lab data was really only good to 2 sig figs. So, if we round both values to 2 sig figs, we would get exactly 41% both times!  Determine the mass percent of PRACTICE WATER in PACKET #46 potassium sulfide pentahydr ate. ANSWER: potassium sulfide pentahydrate = K2S 5H2O 2K = 2 39.1 = 78.2 1S = 1 32.1 = 32.1 5 H2O = 5 18.0 = 90.0 = 200.3 Water is 90.0/200.3 or 44.9% of this hydrate