Combustion and Hydrate Analysis Learning Objectives: - To apply percent composition and empirical formulas to experimental situations.

Agenda Review of Combustion Practice Problem Hydrate Analysis

Some terms... What was the law of definite proportions? What was the law of conservation of mass? During a chemical reaction, the total mass of the substances involved does not change. What was the law of definite proportions? What is percent composition? Elements always combine to form compounds in fixed proportions by mass. (eg. Pure water always contains oxygen and hydrogen combined in the following proportions: 11% hydrogen and 89% oxygen The relative mass of each element in the compound. (Remember, some elements have a bigger atomic mass than others).

A bit of Review... What two kinds of formulas did we learn about? Describe one way to determine the empirical formula.

Combustion Analysis Hydrocarbons are an important and diverse group of compounds Chemists can determine the % composition by combustion analysis Used anywhere unknown compounds need to be analyzed!

How It Works! Hydrocarbons are burned in pure oxygen Carbon dioxide and water are produced CuO oxidizes unreacted C and CO to make CO2 All H becomes H2O and all C becomes CO2 H2O and CO2 are absorbed separately Masses of each are used to determine % composition

Complete Combustion __ CH4 (g) + __ O2 (g)  Incomplete Combustion __ C3H8 (g) + __ O2 (g) 

Sample Problem A 1.00g sample of a pure compound, containing only carbon and hydrogen, was combusted in a carbon-hydrogen combustion analyzer. The combustion produced 0.619g of water and 3.338g of carbon dioxide. a) Calculate the masses of carbon and hydrogen in the sample. b) Find the empirical formula of the compound.

Solutions CxHy (g) + O2 (g)  H2O(g) + CO2(g) Part a) We need to find the mass of carbon and hydrogen in this reaction. How can we tell? All the hydrogen atoms are converted into water. All the carbon atoms are converted into carbon dioxide. If I know the mass of water, and the mass of hydrogen… I can use… CxHy (g) + O2 (g)  H2O(g) + CO2(g)

Percent Composition Percent composition is the percent by mass of each element in a formula. Example: Calculate the percent composition of CO2. CO2 = 1C(12.01g) + 2O(16.00 g) = 44.01 g/mol 12.01 g C x 100 = 27.29% C 44.01 g CO2 32.00 g O x 100 = 72.71% O 44.01 g CO2 100.00 % Basic Chemistry Copyright © 2011 Pearson Education, Inc.

Solutions Find the mass percent (as a decimal) of hydrogen in water. Multiply the mass percent by the mass of water (given in the question). This will give us the mass of hydrogen in the water. Do the same for carbon.

Solutions CxHy (g) + O2 (g)  H2O(g) + CO2(g) Part b) We need to find the empirical formula for the compound We use our rhyme… except we can skip the first step. CxHy (g) + O2 (g)  H2O(g) + CO2(g)

Hydrate Analysis Hydrates are ionic compounds with water incorporated into their crystal structure Anhydrous compounds have no water attached Compare MgSO4 7H2O and MgSO4 H2O

Hydrate Analysis The molar mass and the % composition must include the mass of any water molecules in the compound when doing calculations. Example: LiCl2 · 4H2O The mass of this compound includes the mass of the ionic compound AND water.

Sample Problem A hydrate of barium hydroxide, Ba(OH)2·xH2O, is used to make barium salts and to prepare certain organic compounds. Since it reacts with CO2 from the air to yield barium carbonate, BaCO3, it must be stored in tightly stoppered bottles. A 50g sample of the hydrate contains 27.2g of Ba(OH)2. Calculate the percent, by mass, of water in Ba(OH)2·xH2O. Find the value of x in Ba(OH)2·xH2O.

Solutions a) A 50g sample of the hydrate contains 27.2g of Ba(OH)2. Calculate the percent, by mass, of water in Ba(OH)2·xH2O. Ba(OH)2·xH2O. We can find the percentage of water leftover by finding the mass of water. Total Mass of sample = 50g 27.2g ? g

Solutions Ba(OH)2·xH2O. Convert BOTH parts into moles. b) Find the value of ‘x’ in Ba(OH)2·xH2O. Convert BOTH parts into moles. Ba(OH)2·xH2O. We want to find the amount of water for every 1 Ba(OH)2. Divide the moles of barium hydroxide and water by the moles of barium hydroxide.