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CHEMISTRY 30S – MODULE 3 CHEMICAL REACTIONS LESSON 6 Outcomes When you have completed this lesson, you will be able to: Calculate the percent composition of a compound given its formula. Calculate the empirical formula of a compound from percent composition or mass data. Calculate the molecular formula of a compound from percent composition or mass data. When you get an assignment back, it’s usually out of 10 or 20 or 30, not always 100. But we always convert it to a percentage (out of 100) so we can compare them more easily. We take your mark, say 25 out of 30, divide what you got, by the total and multiply by 100 to get the percentage – 25/30 = 0.833333 and moving the decimal point two places to the right we get 83.3 % Chemists do chemical analysis to find out the composition of a substance. They determine the relative amounts of each element contained in the compound, using mass, and express these amounts as a “percent composition” by mass. The calculation is very similar to that described above for finding your marks on an assignment or test – the mass of oxygen let’s say, in a sample of a compound is divided by the total mass of the sample and then multiplied by 100. Example 1. A compound contains 22.1 g of copper, 11.2 g of sulphur and 22.3 g of oxygen. What is the percent composition of each element in this compound? Solution: Step 1. Determine the total mass of the sample of the compound. Just like with the test, to determine the percent composition we must first know what the total mass of the sample is: 22.1 g + 11.2 g + 22.3 g = 55.6 g Step 2. Determine percent composition of each. Copper: 22.1/55.6 = 0.3977 x 100 = 39.8% Sulphur: 11.2/55.6 = 0.201 x 100 = 20.1% CHEM 30S Mod 3 Chemical Reactions Lesson 6 1 TALC 2010 Oxygen: 22.3 / 55.6 = 0.401 x 100 = 40.1% Percent Composition From Chemical Formulas If the formula of a compound is known, the percent composition can be determined. We can determine the amount of each element in a predetermined amount of the compound, like a mole of that compound. We use the amount of each element in one mole of the compound, divide by the molar mass of the compound and multiply by 100, or % composition = mass of element in 1 mol of compound molar mass of compond Example 2. Determine the percent composition of each element in Na2CO3. Solution. Step 1. Determine the number of moles of each element in one mole of the compound. Recall, the number of moles of each element in one mole of the compound is equal to the subscript for that element. 1 mole of Na2CO3 contains: Na = 2 moles C = 1 mole O = 3 moles Step 2. Determine the mass of each element in one mole of the compound. The mass of each element is the number of moles of that element multiplied by the molar mass of the element. Mass of sodium = 23 x 2 = 46 g mass of carbon = 12 x 1 = 12 g mass of oxygen = 16 . 3 = 48 g Step 3. Find the fraction of each and convert to a percentage. To find the fraction, divide the mass of each element in one mole, by the mass of one mole (molar mass) of the compound. Na2CO3 = 106.0 g/mol Sodium: 46 g/106.0 = 0.4339 x 100 = 43.4% Carbon: 12 g/106 = 0.113 x 100 = 11.3 % Oxygen: 48 g/106 = 0.453 x 100 = 45.3% CHEM 30S Mod 3 Chemical Reactions Lesson 3 2 TALC 2010 Example 3. Determine the percent, by mass, of water in MgSO4·9H2O. Solution. Step 1. Determine the number of moles of water in one mole of the compound. For every mole of magnesium sulphate, there are 9 water molecules Step 2. Determine the mass of each in one mole of the compound. water’s molar mass as 18, so 18 x 9 = 162 g Step 3. Find the fraction of each and convert to a percentage. To find the fraction, divide the mass of each element in one mole, by the molar mass of the compound. MgSO4·9H2O = 282.4 g/mol water makes up 162/282.4 = 57.4% of the mass of the compound. Empirical Formula Once chemists have analyzed a compound and determined its percent composition, they can use this information to determine the compound’s formula. A compound can have two formulas, molecular and empirical formulas. The molecular formula of a compound is the actual formula of the compound. The molecular formula contains the actual number of atoms of each element in one unit of that compound. The empirical formula is the formula of a compound having the smallest whole number ratio of the moles of each element. The subscripts of the molecular formula are always a whole number multiple of the subscripts of the empirical formula. Molecular Formula C2H4, C7H214, C20H40 H2O2, hydrogen peroxide C6H12O6, glucose Na2CO3 CHEM 30S Empirical Formula CH2 HO CH2O Na2CO3 Mod 3 Chemical Reactions Lesson 3 3 TALC 2010 Compound Molecular Formula Empirical Formula Multiplier Formaldehyde CH2O CH2O 1 Acetic acid, vinegar C2H4O2 CH2O 2 Glucose C6H12O6 CH2O 6 To determine the empirical formula of a substance you need to know the elements in the compound and the percent composition of each. Then 1. Assume that the total mass of the sample is 100 g. This means the mass of each element in 100 g of the compound is equal to its percent composition. 2. Determine the number of moles of each element in 100 g of compound. 3. Calculate the lowest whole number ratio by dividing each mole value by the smallest mole value . Example 4. Calculate the empirical formula of a compound whose percent composition is 58.8% carbon, 9.80% hydrogen and 31.4% oxygen. Solution. Step 1. Assume that the total mass of the sample is 100 g. In a 100 g sample, there will be 58.8 g carbon, 9.80 g hydrogen and 31.4 g of oxygen. Step 2. Find the number of moles of each element in the 100-g sample. C: n = m/M n = 58.8 g/12g/mol n = 4.9 mol of C in 100 g of compound H : n = m/M n = 9.8 g/1.01 g/mol n = 9.7 mol of H in 100 g of compound O: n = m/M n = 31.4 g/16 g/mol n = 1.96 mol of O in 100 g of compound Step 3. Now you take the smallest number and divide every other number of moles CHEM 30S Mod 3 Chemical Reactions Lesson 3 4 TALC 2010 by that. C: 4.9/1.96 = 2.5 H: 9.7/1.96 = 4.95 (close enough to 5, we round it) O: 1.96/1.96 = 1 From this the empirical formula appears to be: C2.5H5O1 Since this is not the lowest whole-number ratio, we must multiply each subscript by a value that will give the lowest whole-number ratio (in this case it is 2) 2×(C2.5H5O1) = C(2.5 × 2)H(5 × 2)O(1 × 2)= C5H10O2 The empirical formula is C5H10O2. In class we had practice questions assigned. If you missed class and want to practice, use the following exercise from WebCT: Exercise Answer the following questions. Remember showing all your work is good practice. 1. Determine the percent composition of each element in the following. a) H2SO4 b) Fe(C2H3O2)3 c) C12H22O11 d) aspartame, C14H18N2O5 2. What are the empirical formulas of the following compounds? a) C2H4(OH)2 b) H2S2O8 c) C4H10 d) C2H5OH e) C6H8O6 f) B2H6 3. A sample of Freon, a gas once used as a propellant in aerosol cans, was found to contain 0.423 g of C, 2.50 g Cl, and 1.34 g F. What is the empirical formula of this compound? 4. Determine the empirical formula for each of the following. a) 0.104 mol potassium, 0.052 mol carbon, and 0.156 mol oxygen b) 35.98% aluminum and 64.02% sulfur CHEM 30S Mod 3 Chemical Reactions Lesson 3 5 TALC 2010 c) 22.9% sodium, 21.5% boron, and 55.7% oxygen d) 21.7% carbon, 9.6% oxygen, and 68.7% fluorine 5. A compound that contains only carbon, hydrogen and oxygen is 48.64% carbon and 8.16% hydrogen by mass. What is the empirical formula of this substance? CHEM 30S Mod 3 Chemical Reactions Lesson 3 6