Determining the Chemical Formula of a Substance

Determining the Chemical Formula of a Substance Chemical formulas of substances can be divided into empirical formulas and formulas molecule. The empirical formula can be determined by calculating the moles of the components constituents of substances using molar masses. As for the molecular formula can be determined if the empirical formula and the relative molecular mass (Mr) of the substance is known. a. Determining the Empirical Formula of a Substance In determining the empirical formula, the mole ratios of the elements in substance must be the simplest comparison. Example 1. A number of samples of the substance contained 11.2 grams of Fe and 4.8 grams of O (Ar Fe = 56 and O = 16). Determine the empirical formula of the compound! Answer: To determine the empirical formula of a substance, we calculate the mole ratio of Fe to O as follows. The ratio of Fe : O = 0.2 : 0.3 = 2 : 3 is obtained. So, the empirical formula of the compound is Fe2 O3 . Example 2. Determining Empirical Formulas Based on Mass Percent of Substance Composition Elements Vanilla which is used to flavor food has the following composition: 63.2% C, 5.2% H, and 31.6% O (Ar C = 12, H = 1, and O = 16). Determine the empirical formula! Answer: To determine the empirical formula of vanilla, we calculate the mole ratios of C, H, and O. Suppose in 100 grams of vanilla sample. The mole ratio C : H : O = 5.27 : 5.2 : 1.98 is obtained = 2.66 : 2.66 : 1 =8:8:3 So, the empirical formula of vanilla is C8 H8 O3 . Exercise 1) A compound has a composition of 22.9% Na, 21.5% B, and 55.7% O (ArNa = 23, B = 10, O = 16). Determine the empirical formula! 2) A nitrogen oxide compound consists of 7 grams of nitrogen and 12 grams of oxygen (Ar N = 14 and O = 16). Determine the empirical formula of the nitrogen oxide! 3) Food flavoring ingredient monosodium glutamate (MSG) has a composition of 13.6% Na, 35.5% C, 4.8% H, 8.3% N, and 37.8% O (Ar Na = 23, C = 12, H = 1, N = 14,= 16). Determine the MSG empirical formula! b. Determining the Molecular Formula of a Substance Basically, molecular formulas are multiples of ru empirical mus. As an example: To determine the molecular formula then: with n = integer The value of n can be determined if the empirical formula and relative molecular mass (Mr ) of the substance are known. Example 1. A compound with the empirical formula CH (Ar C = 12 and H = 1) has Mr = 26. Determine the molecular formula of the compound! Answer: Mr = n × (Ar C + Ar H) 26 = n × (12 + 1) 26 = n × 13 n=2 So, the molecular formula of the compound is (CH)2 = C2 H2 . Exercise 1. A substance has a Mr of 181.5. If the empirical formula is C2 HCl (Ar C = 12, H = 1, Cl = 35.5), determine the molecular formula! 2. Determine the empirical formula of a compound containing: a. 26.53% K, 35.37% Cr, and the rest is oxygen b. 29.11% Na, 40.51% S, and the remainder oxygen (Ar K = 39, Cr = 52, O = 16, Na = 23, and S = 32) 3. In complete combustion 2.3 grams of a compound containing C, H, and O 4.4 grams of CO2 and 2.7 grams of H2 O are produced. The equation for the reaction is: Cx Hy Oz + O2 ÿÿÿ CO2 + H2 O Determine the empirical formula of the compound! (Ar C = 12, H = 1, and O = 16) 4. In the complete combustion of 13 grams of a Cx Hy hydrocarbon , 4.4 grams of CO2 is produced . The mass of 5 liters of compound (T, P) is 6.5 grams. At the same (T, P), the mass of 1 liter of oxygen is 1.6 grams. Determine the molecular formula of the hydrocarbon! (Ar C = 12, H = 1, and O = 16) 5. A gaseous Cx Hy hydrocarbon consists of 80% carbon and the remainder hydrogen. Determine the empirical formula of the compound. If it is known that the mass of 1 liter of the compound (STP) is 1.34 grams, determine the molecular formula of the hydrocarbon compound! 6. The Cx Hy Oz compound is composed of 40% carbon, 6.67% hydrogen, and the remainder oxygen. If the Mr of the compound is 90, determine the molecular formula of the compound!