90.14 g/mol x g/mol. Molecular formula: molecular formula 2 empirical formula 2 C OH C O H

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1 Whole-number multiple: M x M actual compound C2OH g/mol x g/mol g/mol x g/mol 2 Molecular formula: molecular formula 2 empirical formula 2 C OH 2 5 C O H Check Your Solution Work backward by calculating the molar mass of C 4 O 2 H 10 (). Molar mass, M, of C 4 O 2 H 10 (): M 4 M 2 M 10M C4O2H10 C O H g/mol g/mol g/mol g/mol The calculated molar mass matches the molar mass that is given in the problem. 44. Practice Problem (page 275) The empirical formula for styrene is CH(), and its molar mass is 104 g/mol. What is its molecular formula? What Is Required You need to determine the molecular formula for styrene. What Is Given? You know the empirical formula: CH You know the molar mass of the actual compound: 104 g/mol Plan Your Strategy Determine the molar mass of CH(). To find the whole-number multiple, x, divide the experimentally determined molar mass by the molar mass of the empirical formula Chapter 6ProportionsinChemicalCompounds MHR 77

2 To determine the molecular formula, multiply each subscript of the empirical formula by the whole-number multiple. Act on Your Strategy Molar mass, M, of CH(): M 1 M 1M CH C H g/mol g/mol g/mol Whole-number multiple: M x M actual compound CH 104 g/mol x g/mol 104 g/mol x g/mol Molecular formula: molecular formula 8 empirical formula 8 CH C H 8 8 Check Your Solution Work backward by calculating the molar mass of C 8 H 8 (): Molar mass, M, of C 8 H 8 (): M 8 M 8M C8H8 C H g/mol g/mol g/mol The calculated molar mass closely matches the molar mass that is given in the problem Chapter 6ProportionsinChemicalCompounds MHR 78

3 g/mol mass percent of H g/mol mass percent of N % 10.90% g/mol g/mol % 18.90% g/mol mass percent of O g/mol % 21.59% The compound is 48.61% carbon, 10.90% hydrogen, 18.90% nitrogen, and 21.59% oxygen. These percentages closely match the given percentage composition. 50. Practice Problem (page 276) Estradiol is the main estrogen compound that is found in humans. Its molar mass is g/mol. The percentage composition of estradiol is 72.94% carbon, 10.80% oxygen, and 8.16% hydrogen. Determine whether its molecular formula is the same as its empirical formula. If not, what is each formula? What Is Required? You need to find the empirical and molecular formulas for estradiol. What Is Given? You know the composition of the compound: 72.94% carbon, 10.80% oxygen, 8.16% hydrogen Plan Your Strategy Assume that the mass of the sample is g. Determine the molar masses of C, H, and O using the periodic table. Convert each mass to amount in moles. Determine the ratio of amounts of each element in whole numbers by dividing each mole amount by the lowest mole amount Chapter 6ProportionsinChemicalCompounds MHR 90

4 Write the empirical formula for the compound. Determine the empirical formula molar mass. To find the whole-number multiple, x, divide the given molar mass by the empirical formula molar mass. To determine the molecular formula, multiply each subscript of the empirical formula by the whole-number multiple. Act on Your Strategy In a g sample of the compound, there will be g of C, 8.16 g of H, and g of O. From the periodic table: The molar mass of carbon is g/mol. The molar mass of hydrogen is 1.01 g/mol. The molar mass of oxygen is g/mol. Amount in moles, n, of each element: g nc g /mol mol 8.16 g nh 1.01 g /mol mol no g /mol g mol Whole-number ratio: : : Ratio: :11.968:1 = 9:12:1 The empirical formula for the compound is C 9 H 12 O Chapter 6ProportionsinChemicalCompounds MHR 91

5 Molar mass, M, of C 9 H 12 O: M 9 M 12 M 1M C9H12O C H O g/mol 12(1.01 g/mol) + 1(16.00 g/mol) g/mol Whole-number multiple: M x M actual compound C9H12O g/mol x g/mol g/mol x g/mol Molecular formula: molecular formula 2 empirical formula 2 C H O 9 12 C H O Check Your Solution Work backward. Determine the percentage composition of C 18 H 24 O 2 : Molar mass, M, of C 18 H 24 O 2 : M 18 M 24 M 2M C18H 24O2 C H O g/mol 24(1.01 g/mol) + 2(16.00 g/mol) g/mol g/mol mass percent of C g/mol % 79.36% g/mol mass percent of H g/mol % 8.89% Chapter 6ProportionsinChemicalCompounds MHR 92

6 g/mol mass percent of O g/mol % 11.75% The compound is 79.36% carbon, 8.89% hydrogen, and 11.75% oxygen. The percentages add to 100.0%. These percentages are in the range of the given percentage composition. Section 6.2 Empirical and Molecular Formulas Solutions for Practice Problems Student Edition pages Practice Problem (page 278) Calculate the percent by mass of water in the compound MgSO 3 6H 2 O(s). What Is Required? You need to calculate the percent by mass of water in the hydrate MgSO 3 6H 2 O(s). What Is Given? You know the chemical formula for the sample: MgSO 3 6H 2 O Plan Your Strategy Determine the molar mass of MgSO 3 6H 2 O(s). Calculate the percent by mass of water. Act on Your Strategy Molar mass, M, of MgSO 3 6H 2 O(s): M 1 M 1 M 3M 6M MgSO 3 6H2O Mg S O H2O g/mol + 1(32.07 g/mol g/mol + 6[ g/mol g/mol ] g/mol + 1(32.07 g/mol) g/mol + 6(18.02 g/mol) g/mol Chapter 6ProportionsinChemicalCompounds MHR 93

NOTES: 10.3 Empirical and Molecular Formulas

NOTES: 10.3 Empirical and Molecular Formulas What Could It Be? Empirical Formulas Indicate the lowest whole number ratio of the atoms in a compound: 1) Determine moles of each element present in the compound