8.8 Percentage Composition
The proportion by mass of elements in a given compound is always the same. Water always contains 11% H and 89% O. Therefore, the percentage composition of water is 11% H and 89% O. The percentage composition of a compound is the percentage (by mass) of each of the elements in the compound.
Example: (from pg 190) A sample of a compound containing carbon and oxygen had a mass of 88g. Experimental procedures showed that 24g of this sample with carbon, and the remaining 64g was oxygen. What is the percentage composition of this compound?
% carbon = (mass of carbon / mass of sample) x 100% = (24g / 88g) x 100% = 27%
% oxygen = (mass of oxygen / mass of sample) x 100% = (64g / 88g) x 100% = 73%
The percentage composition of the compound is 27% carbon and 73% oxygen
8.9 Determining the Formula of a Compound
You can find the empirical if you know: (1) Mass of each element in a sample of the compound or (2) the percentage composition of the compound.
Example:(#12 on pg 192) Laboratory procedures show that a 9.2g sample of a compound is 2.8g nitrogen and 6.4g oxygen. Find the empirical formula of the compound.
Step 1: Convert the gram masses to moles using the correct conversion factors.
2.8g N x (1.0 mol N atoms / 14 g N)= .20 mol N atoms
6.4g O x (1.0 mol O atoms / 16 g O)= .40 mol O atoms
Since there are 2 times as many moles of O as N, that means that there are twice as many oxygen atoms in the compound as there are nitrogen atoms.
Step 2: Using the number of moles, determine the smallest whole number ratio for O and N atoms.
(Number of moles of O atoms / Number of moles of N atoms) = (.40 / .20) = 2/1
Step 3: Use the numbers from the ratio (step 2) as subscripts in the empirical formula: N1O2 (the 1 can be omitted).
8.10 Another Way to Determine Empirical Formulas
The empirical formula of a substance can also be found by using the number of moles of each element in the compound.
Example: A sample of a compound containing carbon and hydrogen is decomposed to make .0134g of solid carbon and .0500 dm3 at standard temp and pressure of gaseous H2. What is the empirical formula of the compound?
Step 1: Find the number of moles of all the elements mentioned.
.0134g C x (1.00 mol C / 12.0g C) = .00112 mol C
.0500 dm3 H2 x (1.00 mol H2 / 22.4 dm3 H2 ) = .00223 mol H2
Step 2: As in the previous example, find the smallest whole-number ratio for the moles of C to moles of H. Since .00112 is the smaller number, divide the number of moles for C and H2 by the moles of C.
(.00112 mol C / .00112) = 1.00 mol C
(.00223 mol H2 / .00112) = 2.00 mol H2 = 4.00 mol H
Because H2 is being produced as the diatomic gas, you must multiply the moles of H2 by 2 to find the moles of H. Therefore, the empirical formula is CH4