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Chapter 10 Chemical Quantities

Chapter 10 Chemical Quantities. 10.3 Calculating Percent Composition of a Substance. Chemistry. Today we are learning to:- 1. Describe what is meant by water of crystallization (section 15.2) 2. Calculate the percentage composition of a compound

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Chapter 10 Chemical Quantities

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  1. Chapter 10 Chemical Quantities 10.3 Calculating Percent Composition of a Substance

  2. Chemistry Today we are learning to:-1. Describe what is meant by water of crystallization(section 15.2) 2. Calculate the percentage composition of a compound 3. Calculate the empirical formula of a compound

  3. Exothermic and Endothermic Processes 17.1 Water of Crystallization

  4. Hydrates • An ionic substance will often contain a definite proportion of water as part of its crystal structure. • The water molecules are shown as part of the formula unit. • Ionic substances containing water are called hydrates. • Water of crystalizationcan be removed by heating. The ionic compound is then called anhydrous. • By weighing a hydrated compound before and after heating it is possible to calculate percentage of water present in the hydrated salt and from this, the number of water molecules surrounding each formula unit, by using the percent composition formula. Hydrated copper sulfate: CuSO4 · 5H2O Anhydrous copper sulfate: CuSO4

  5. Percentage Composition • Formula units and molecular formulas represent the composition of a substance {Ca(NO3)2 formula unit for calcium nitrate (ionic) or C2H5OH  molecular formula for Ethanol (covalent)} • The relative amounts of the elements in a compound are expressed as the percent compositionor the percent by mass of each element in the compound.

  6. Exothermic and Endothermic Processes 17.1 Percentage Composition • Two methods: • Calculate the percent composition of each element from a formula. • Calculate the percent composition from experimental data. Ex.1. Calculate the percentage composition of propane C3H8? Step 1 Calculate the gram formula mass of = C3H8 For C  3 x 12.0g = 36.0g For H  8 x 1.0g = 8.0g__ formula mass =44.0g Step 2 Use the formula to calculate % mass of elements: % mass of C = 36.0g x 100 = 81.8% 44.0g % mass of H = 8.0g x 100 = 18.2% 44.0g Note that if your calculations are correct the totals of all of your percentages will add up to 100%

  7. Exothermic and Endothermic Processes 17.1 Percentage Composition • Two methods: • Calculate the percent composition of each element from a formula. • Calculate the percent composition from experimental data. Ex.2. 13.6g of a compound containing only magnesium and oxygen is decomposed and 5.5g of oxygen is obtained. Calculate the percentage composition of the elements in the compound? Step 1 Calculate the mass of magnesium Compound mass = 13.60g Oxygen mass = 5.40g__ Magnesium mass = 8.20g Step 2 Use the formula to calculate % mass of elements: % mass of Mg = 8.20g x 100 = 60.3% 13.60g % mass of O = 5.40g x 100 = 39.7% 13.60g Note that if your calculations are correct the totals of all of your percentages will add up to 100%

  8. Exothermic and Endothermic Processes 17.1 Empirical Formulas • What is an Empirical formula and Why Use It • The empirical formula of a compound shows the smallest whole number ratio of different atoms in a compound. • Empirical formulas are a halfway step in working out the actual molecular formula of a compound. • It is found by first working out the percentage composition by mass of elements in a compound by experiment, as in the example of magnesium oxide. • It may or may not be the same as the molecular formula.

  9. Exothermic and Endothermic Processes 17.1 Empirical Formulas • What is an Empirical formula and Why Use It • You can calculate the molecular formula from the empirical formula if you know the compound’s molar mass/gram formula mass.

  10. Exothermic and Endothermic Processes 17.1 Empirical Formulas • What is an Empirical formula and Why Use It • You can calculate the molecular formula from the empirical formula if you know the compound’s molar mass/gram formula mass.

  11. Exothermic and Endothermic Processes 17.1 Empirical Formulas Ex. A compound is analysed and found to contain 25.9% nitrogen and 74.1% oxygen. What is its empirical formula? The percent composition here gives a ratio of masses. We can change the ratio of masses to a ratio of moles then reduce this new ratio to the lowest whole number ratio. Step 1 %N = 25.9% Empirical formula = NxOy %O = 74.1% Percent means parts per 100. So in 100g of our substance we have25.9g nitrogen and 74.1g oxygen. Step 2 14g N = 1mol 1g N = 1 mol 14g 25.9g N = 25.9g x 1 mol = 1.85mol 14g Step 2 16g O = 1mol 1g O = 1 mol 16g 74.1g O = 74.1g x 1 mol = 4.63mol 14g We need whole number ratios so divide both numbers by the smallest number, then multiply by the smallest whole number that will convert both to whole numbers Step 3 Mole ratio = N1.85O4.63= N1O4.63 = N1O2.5 = N2O5 1.85 Divide both by 1.85 Multiply both by 2

  12. Exothermic and Endothermic Processes 17.1 Empirical Formulas • Calculate the empirical formula for each compound • 94.1% O and 5.9% H • 67.6% Hg, 10.8% S and 21.6% O

  13. END OF SHOW

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