Empirical Formula & Molecular Formula

Empirical Formulas Best way to identify an unknown compound is to determine its chemical formula. Empirical Formula: the smallest whole number ratio of a compound. (Ionic compounds are already in its empirical formulas)

Empirical Vs. Molecular Formulas Empirical Formula a formula that shows the simplest whole-number ratio of elements in compounds Molecular Formula a formula that shows the element symbols & exact number of each type of atom in a molecular compound

can be reduced to lowest ratio can be reduced to lowest ratio CH4 Vs. C6H24 Vs. C18H72 Lowest ratio can be reduced to lowest ratio can be reduced to lowest ratio In empirical & molecular formula 18 can go into 18 & 72--> CH4 6 can go into 6 & 24--> CH4 same empirical formula. But not the same molecular formula

M= 12.01g/mol + (4 x 1.01g/mol) = 16.00 g/mol If you know the chemical formula you can find percent composition. CH4 C6H24 M= 12.01g/mol + (4 x 1.01g/mol) = 16.00 g/mol MC6H24= 96 g/mol MC= 72 g/mol MC= 12.01g/mol MH= 1.01g/mol (72 g/mol ) x 1mol x 100% = 75% 96.00g (12.00g/mol ) x 1mol x 100% = 75% CH4 & C6H24 16.00g - same empirical formula - should be ratios of each other -helps determine percent composition (4.01g/mol) x 1mol x 100% = 25% 16.00g

Determining Empirical Formula Find the empirical formula of a compound with percentage composition 35.4% Na and the remainder nitrogen Step 1: Assume your conveniently working with a mass of 100.0 g. Na= 35.4 g N= 64.6 g 100.0g - 35.4g = 64.6 g Step 2: Calculate the amount of each element. nNa = (35.4 g) (1 mol) nN = (64.5 g) (1 mol) 22.99g 14.01g = 1.540 mol = 4.6110 mol

Determining Empirical Formula Find the empirical formula of a compound with percentage composition 35.4% Na and the remainder nitrogen Step 3: To determine the simplest ratio of the elements in the compound, divide the amount of each element by the smallest amount nNa nN 4.6110 mol 1.540 mol = = nNa nNa 1.540 mol 1.540 mol = 1 = 2.98

Determining Empirical Formula Find the empirical formula of a compound with percentage composition 35.4% Na and the remainder nitrogen Step 2: Calculate the amount of each element. nNa nNa 1.540 mol 4.6110 mol = = nNa nN 1.540 mol 1.540 mol = 1 = 2.98 The ratio for Na & N is1:2.98. Since we cannot have a fraction of an element in a compound, the value of N is rounded off to the nearest whole number. As a result, the simplest whole-number ratio is 1:3........NaN3

If a number is within 0.05 of a whole number, you can round it up or down to the nearest whole number. But what do you do if one of the values is not close to a whole number? You multiply all the subscripts by the fraction denominator gives whole numbers C1 x 3 H1.333 x 3 = C3H4

Practice nc (52.2 g) (1 mol) = = 4.3464 mol 12.01g nH (6.15g) (1 mol) Determine the empirical formula of a compound listed that contains 52.2% carbon, 6.15% hydrogen and 41.7% oxygen Step 1: Assume your conveniently working with a mass of 100.0 g. nc (52.2 g) (1 mol) = = 4.3464 mol 12.01g nH (6.15g) (1 mol) = = 6.0891mol 1.01g n0 (41.7g) (1 mol) = = 2.6063mol 16.00g

Practice nc 4.3464 mol = = 1.67 n0 2.6063mol 6.0891mol nH = = 2.29 n0 Determine the empirical formula of a compound listed that contains 52.2% carbon, 6.15% hydrogen and 41.7% oxygen Step 2: Divide the amount of each element by the smallest amount. nc 4.3464 mol = = 1.67 n0 2.6063mol Step 3: These calculations give an empirical formula of C1.67H2.29O1 Multiplying each of the subscripts by 3 give C5H7O3 6.0891mol nH = = 2.29 n0 2.6063mol n0 2.6063mol = = 1 n0 2.6063mol

Practice Write the empirical formula of compounds with the following percentage 20.2% Al; 79.8% Cl

Practice Write the empirical formula of compounds with the following percentage 50.85% carbon; 8.47% hydrogen; and 40.68% oxygen

Molecular Formula gives the exact number of atoms of each element present “chemical formula” = molecular formula A compound’s molar mass is always a whole number multiple of the molar mass of the empirical formula. You ca find this multiple x, by dividing the molar mass of the compound by the molar mass of the empirical formula

Molecular Formula x = molar mass of compound molar mass of empirical

Find a Molecular Formula Given its Empirical formula Compound with empirical formula CH2 & the molar mass of 84.18 g/mol Step 1: Find the empirical molar mass by adding the molar mass of each element MCH2 = 1( 12.01 g/mol) + 1(1.01g/mol) = 14.03 g/mol

Find a Molecular Formula Given its Empirical formula Compound with empirical formula CH2 & the molar mass of 84.18g/mol Step 2: Solve for x, x = 84.18 g/mol 14.03 g/mol x = 6

Find a Molecular Formula Given its Empirical formula Compound with empirical formula CH2 & the molar mass of 84.18g/mol Step 3: therefore, the molar mass of the compound is 6 times the molar mass of the empirical formula CH2= C6H12 molecular formula

Practice Determine the molecular formulas KSO4; 270.32 g/mol

Using Percentage Composition & Molar Mass Data Determine the molecular formula of vitamin C (ascorbic acid). This compound contains 40.5% Carbon, 4.6% hydrogen, 54.5% oxygen. Its molar mass is 176.14g/mol