 # Percent Comp. Percentage composition Indicates the relative amount of each element present in a compound.

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Percent Comp

Percentage composition Indicates the relative amount of each element present in a compound.

Calculating percentage composition Step 1 : Calculate molar mass Step 2 : Divide the subtotal for each element’s mass by the molar mass. Step 3: Multiply by 100 to convert to a percentage.

Example 1 1. Calculate the molar mass of water, H 2 O. Step 1 – H- 2(1.01)= 2.02 O – 1(16.0)= 16.0 ________ 18.02

Step 2 2.02/18.02 X 100 = 11.21% 16.0 /18.02 X 100 = 88.79 % Water is composed of 11.21 % hydrogen and 88.79 % oxygen.

Example 2 Calculate the percentage composition of sucrose, C 12 H 22 O 11.

Example 3 Find the percentage composition of hydrogen in sulfuric acid.

Empirical Formula From percentage to formula

Empirical Formula The empirical formula gives the ratio of the number of atoms of each element in a compound. CompoundFormulaEmpirical Formula Hydrogen peroxideH 2 O 2 OH BenzeneC 6 H 6 CH EthyleneC 2 H 4 CH 2 PropaneC 3 H 8 C 3 H 8

Calculating Empirical Pretend that you have a 100 gram sample of the compound. That is, change the % to grams. Convert the grams to mols for each element. Write the number of mols as a subscript in a chemical formula. Divide each number by the least number. Multiply the result to get rid of any fractions.

EXAMPLE 1 A compound was analyzed to be 82.67% carbon and 17.33% hydrogen by mass. What is the empirical formula for the compound? Assume 100 g of sample, then 82.67 g are C and 17.33 g are H

EXAMPLE 1 Convert masses to moles: 82.67 g C x mole/12.011 g = 6.88 moles C 17.33 g H x mole/1.008 g = 17.19 mole H Find relative # of moles (divide by smallest number)

EXAMPLE 1 Convert moles to ratios: 6.88/6.88 = 1 C 17.19/6.88 = 2.50 H Or 2 carbons for every 5 hydrogens C 2 H 5

Example 2 Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. Assume 100 g so 38.67 g C x 1mol C = 3.220 mole C 12.01 gC 16.22 g H x 1mol H = 16.09 mole H 1.01 gH 45.11 g N x 1mol N = 3.219 mole N 14.01 gN

Example 2 3.220 mole C 16.09 mole H 3.219 mole N C 3.22 H 16.09 N 3.219 If we divide all of these by the smallest one It will give us the empirical formula

Example 2 The ratio is 3.220 mol C = 1 mol C 3.219 molN 1 mol N The ratio is 16.09 mol H = 5 mol H 3.219 molN 1 mol N C 1 H 5 N 1 is the empirical formula

Molecular Formula Is the actual Formula Only applies to covalent compounds Find the Mass of the Empirical Formula Divide the actual molar mass by the mass of one mole of the empirical formula. Caffeine has a molar mass of 194 g. what is its molecular formula? Empirical Formula C 4 H 5 N 2 O 1

Caffeine Example Find the mass of the empirical formula Empirical Formula C 4 H 5 N 2 O 1 C= 4 X 12.01 H = 5 X 1.01 N = 2 X 14.01 O = 1 X 15.99 Total Mass = 97g

Caffeine Example Find x if 194 g 97 g = 2 C4H5N2O1C4H5N2O1 C 8 H 10 N 4 O 2. 2 X

Example 2 A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula?

Example 2 71.65 Cl 24.27C 4.07 H. = 2.0mol = 4.0mol

Example 2 Cl 2 C 2 H 4 Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula? We divide by lowest (2mol ) Cl 1 C 1 H 2 would give an empirical wt of 48.5g/mol

Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula? would give an empirical wt of 48.5g/mol = 2 =

2 X Cl 1 C 1 H 2 = Cl 2 C 2 H 4

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