Pick up notes Get out your periodic table

Pick up notes. Get out your periodic table and calculator. Molar Quantities lab due Tuesday.

The simplest whole number ratio of elements in a compound is called the EMPIRICAL FORMULA. In contrast, the molecular formula of a compound is the actual number of atoms of each element in the compound. It can be calculated from the mole ratio or from percent composition data.

Molecular Formula C 6 H 12 O 6 CO P 4 O 10 Empirical Formula CH 2 O CO P 2 O 5

What is the empirical formula for a compound if a 2. 50 g sample contains 0. 900 g of calcium and 1. 60 g of chlorine? Step One: Determine the number of moles of Ca and Cl. (You must first determine the atomic mass of each. ) Ca 0. 900 g Ca Cl 1. 60 g Cl 1 mole Ca = 0. 0225 mol 40. 08 g Ca 1 mole Cl 35. 45 g Cl = 0. 0451 mol

Step Two: Obtain the smallest ratio for the set of moles. Divide both numbers of moles by the smaller number of moles. The smallest is 0. 0225 mol. Ca 0. 0225 mol = 1. 00 0. 0225 mol Cl 0. 0451 mol = 2. 00 0. 0225 mol These should be close to whole numbers MOST of the time.

Step Three: Look at the ratio of the two numbers. Round off to whole numbers (there are exceptions to this). Ca 1. 00 becomes Ca 1 Cl 2. 00 becomes Cl 2 The empirical formula is Ca. Cl 2

1. 2. 3. Convert mass to mole for each element given. Divide the smallest mole into itself and all others. Determine the empirical formula.

A. Find the empirical formula of the compound with 63. 0 g Rb and 5. 90 g O.

Sometimes you are given percentages, not masses, and asked to determine the empirical formula. If this is the case, you can change the percentages directly to grams because you assume a 100 g sample (100%)

A compound has a percentage composition of 40. 0% C, 6. 71% H and 53. 3% O. What is the empirical formula? Assume that there is 100 g (from 100%). In a 100 g sample, there would be 40. 0 g C, 6. 71 g H, and 53. 3 g O. Then change the quantities to moles.

C 40. 0 g C 1 mole C = 3. 33 mol C 12. 01 g C H 6. 71 g H 1 mole H = 6. 64 mol H 1. 01 g H O 53. 3 g O 1 mole O = 3. 33 mol O 16. 00 g O

To obtain the smallest ratio, divide both numbers of moles by the smaller number of moles (3. 33 mol). C 3. 33 mol = 1. 00 3. 33 mol O 3. 33 mol = 1. 00 3. 33 mol H 6. 64 mol = 1. 99 3. 33 mol

Look at the ratio of the two numbers. Round off to whole numbers. C H O 1. 00 becomes 1. 99 becomes 1. 00 becomes C H O The empirical formula is CH 2 O 1 2 1

A. What is the empirical formula for a compound that has 32. 8% Cr and 67. 2% Cl?

What is the empirical formula of a compound that contains 53. 73% Fe and 46. 27% S?

If the ratio is: X. 1 or X. 9, round up or down If the ratio is X. 2 to X. 8, determine what to multiply it by to get a whole number. Examples: X. 3 X. 5 x 3 = X. 9 which can be rounded x 2 = X. 0 which is a whole number X. 4 x 5 = X. 0 which is a whole number

Percent Composition and Empirical Formula Practice handout is due tomorrow. Molar Quantities lab is due Tuesday.