# Empirical and Molecular Formulas Part 2: Calculations using percent composition.

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Empirical and Molecular Formulas Part 2: Calculations using percent composition

Introduction Chemical formulas that have... the lowest whole number ratio of atoms in the compound are empirical formulas. the total number of atoms in the compound are molecular formulas.

Introduction If we know the chemical formula of a compound, then we can find the percent composition of each of the atoms in the compound. We use the molar mass of the compound and the average atomic masses of the atoms in the compound. percent composition = ×100% atomic mass of atoms molar mass of compound

Finding the Empirical Formula If we know the percent composition of a compound, then we can find the empirical formula of the compound. First, we assume that we have 100 g of the compound and find the mass of each atom in the compound. Second, we find the number of mols of each atom. Third, we find the lowest whole number ratio of mols.

Finding the Empirical Formula For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O. Step 1: Assume 100 g of compound 52.2 g C, 13.1 g H, and 34.7 g O Step 2: Find the number of moles of each atom 52.2 g nC=nC= mCmC MCMC = 12.0 g/mol = 4.35 mol

Finding the Empirical Formula For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O. Step 1: Assume 100 g of compound 52.2 g C, 13.1 g H, and 34.7 g O Step 2: Find the number of moles of each atom 13.1 g nC=nC= 4.35 mol 34.7 g52.2 g

Finding the Empirical Formula For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O. Step 1: Assume 100 g of compound 52.2 g C, 13.1 g H, and 34.7 g O Step 2: Find the number of moles of each atom 13.1 g nC=nC= 4.35 mol 34.7 g52.2 g nH=nH= mHmH MHMH = 1.01 g/mol = 13.0 mol

Finding the Empirical Formula For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O. Step 1: Assume 100 g of compound 52.2 g C, 13.1 g H, and 34.7 g O Step 2: Find the number of moles of each atom 13.1 g nC=nC= 4.35 mol 34.7 g52.2 g nH=nH= 13.0 mol

Finding the Empirical Formula For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O. Step 1: Assume 100 g of compound 52.2 g C, 13.1 g H, and 34.7 g O Step 2: Find the number of moles of each atom 13.1 g nC=nC= 4.35 mol 34.7 g52.2 g nH=nH= 13.0 mol nO=nO= mOmO MOMO = 16.0 g/mol = 2.17 mol

Finding the Empirical Formula For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O. Step 1: Assume 100 g of compound 52.2 g C, 13.1 g H, and 34.7 g O Step 2: Find the number of moles of each atom 13.1 g nC=nC= 4.35 mol 34.7 g52.2 g nH=nH= 13.0 mol nO=nO= 2.17 mol

nOnO nOnO Finding the Empirical Formula For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O. Step 1: Assume 100 g of compound 52.2 g C, 13.1 g H, and 34.7 g O Step 2: Find the number of moles of each atom 13.1 g nC=nC= 4.35 mol 34.7 g52.2 g nH=nH= 13.0 mol nO=nO= 2.17 mol Step 3: Find the lowest whole number ratio of mols *Remember, the element with the lowest number of mols goes in the denominator. nCnC 4.35 mol nHnH 13.0 mol 2.17 mol nCnC == 2 1 == 6 1 nOnO This gives an empirical formula of: C H O 2 6

Finding the Empirical Formula For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O. Step 1: Assume 100 g of compound 44.9 g K, 18.4 g S, and 36.7 g O Step 2: Find the number of moles of each atom 44.9 g nK=nK= mKmK MKMK = 39.1 g/mol = 1.15 mol

Finding the Empirical Formula Step 1: Assume 100 g of compound 44.9 g K, 18.4 g S, and 36.7 g O Step 2: Find the number of moles of each atom 18.4 g nK=nK= 1.15 mol 36.7 g44.9 g For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O.

Finding the Empirical Formula Step 1: Assume 100 g of compound 44.9 g K, 18.4 g S, and 36.7 g O Step 2: Find the number of moles of each atom 18.4 g nK=nK= 1.15 mol nS=nS= mSmS MSMS = 32.1 g/mol = 0.573 mol For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O.

Finding the Empirical Formula Step 1: Assume 100 g of compound 44.9 g K, 18.4 g S, and 36.7 g O Step 2: Find the number of moles of each atom nK=nK= 1.15 mol nS=nS= 0.573 mol For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O.

Finding the Empirical Formula Step 1: Assume 100 g of compound 44.9 g K, 18.4 g S, and 36.7 g O Step 2: Find the number of moles of each atom nK=nK= 1.15 mol 36.7 g nS=nS= 0.573 mol nO=nO= mOmO MOMO = 16.0 g/mol = 2.29 mol For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O.

Finding the Empirical Formula Step 1: Assume 100 g of compound 44.9 g K, 18.7 g S, and 36.7 g O Step 2: Find the number of moles of each atom nC=nC= 4.35 mol nH=nH= 13.0 mol nO=nO= 2.29 mol For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O.

K S O 4 nK=nK= 1.15 mol nO=nO= 2.29 mol n S = 0.573 mol nSnS nKnK 1.15 mol nOnO nOnO 2.29 mol Finding the Empirical Formula Step 1: Assume 100 g of compound 44.9 g K, 18.7 g S, and 36.7 g O Step 2: Find the number of moles of each atom Step 3: Find the lowest whole number ratio of mols *Remember, the element with the lowest number of mols goes in the denominator. nCnC == 2 1 == 1 This gives an empirical formula of: 2 For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O. nSnS 0.573 mol 4 = K2SO4K2SO4

Summary There is a three step process to finding the empirical mass of a compound when we know the percent composition. First, we assume that we have 100 g of the compound and find the mass of each atom in the compound. Second, we find the number of mols of each atom. Third, we find the lowest whole number ratio of mols.

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