Determining Chemical Formulas: Empirical and Molecular Formulas, Study Guides, Projects, Research of Chemistry

Download Determining Chemical Formulas: Empirical and Molecular Formulas and more Study Guides, Projects, Research Chemistry in PDF only on Docsity! Chapter 7 Section 4: Determining Chemical Formulas Objective 1: Define empirical formula, and explain how the term applies to ionic and molecular compounds. Objective 3: Explain the relationship between the empirical formula and the molecular formula of a given compound. Empirical Formula: The simplest formula that represents the whole number ratio between the elements in a compound. For ionic compounds the formula is the lowest ratio of the atoms; therefore the empirical formula and the chemical formula are the same – caution is needed when considering polyatomic ions. They are units that cannot be reduced: example peroxide O2 -2 with H it looks like it should be reduced. H2O2  HO, is not correct, the peroxide polyatomic ion is not represented. Molecular formulas are different. Molecular Formula: a formula showing the types and numbers of atoms combined in a single molecule of a molecular compound Molecular formulas and empirical formulas are not always the same. For example, the empirical formula of the gas diborane is BH3, but the molecular formula is B2H6. In this case, the number of atoms given by the molecular formula corresponds to the empirical ratio multiplied by two. Subscripts in a chemical formula are usually thought of as a ratio of atoms. EX: H2O is a ratio of 2 H : 1 O. Subscripts are also a ratio of moles. To determine an empirical formula, one must determine the mole ratio. Objective 2: Determine an empirical formula from either a percentage or a mass composition. Determining the Empirical Formula (Mole Ratio): Using a percent composition – employ 3 steps: 1. Find the number of moles of each element present. 2. Determine the whole number mole ratio. 3. Use the mole ratio for the subscripts of each element in the formula. Sample determination: Determine the empirical formula of a compound that is composed of 36.5% sodium, 25.4% sulfur, and 38.1% oxygen. 1. Find the number of moles of each element present. Since the amount of each element is given in percentage, you must convert the percentage to a mass. If you have 100 grams of the sample, the percentages given are the same as grams. Then calculate the number of moles from this value for each element present. 36.5% Na  36.5g Na 25.4% S  25.4g S 38.1% O  38.1g O Convert that mass to a mole using molar mass. We cannot use decimal values for subscripts, so on to step 2 2. Determine the whole number mole ratio. Divide each mole number by the smallest mole number. This will give a mole to mole ratio for each element in the compound. Ratios: Na = 1.59/0.79 = 2 (2.012) S = .79/.79 = 1 O = 2.38/.79 = 3 (3.012) (36.5g Na) 1 mol Na = 1.59 mol Na 22.99 g Na (25.4g S) 1 mol S = 0.79 mol S 32.07 g S (38.1g O) 1 mol O = 2.38 mol O 16.00 g O Sample Problem Analysis of a 10,150 g sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of 4.433 g. What is the empirical formula of this compound? Solution 1 Analyze Given: sample mass = 10.150 g phosphorus mass = 4.433 g Unknown: empirical formula 2 PLAN Mass composition —> composition in moles ——> smallest whole-number ratio of atoms 3 COMPUTE ie mass of oxygen is found by subtracting the phosphorus mass from the sample mass. sample mass — phosphorus mass = 10.150 g- 4.433 g= 5.717 g Mass composition: 4.433 g P, 5.717 gO Composition in moles: 4433 gp EP My 1431 mol P ! x ———— =H). mo. “30.97. ¢P- i mol O 5.717 gO x ——— = 0.3573 mol O 16.00 ¢0 Smallest whole-number mole ratio of atoms: 0.1431 mol P 0.3573 mol O 0.1431 7° 0.1431 1 mol P : 2.497 mol O The number of O atoms is not close to a whole number. But if we multiply each number in the ratio by 2, then the number of O atoms becomes 4.994 mol, which is close to 5 mol. The simplest whole-number mole ratio of P atoms to O atoms is 2:5. The compound's empirical formula is P,Q. 4 EVALUATE The arithmetic is correct, significant figures have been used correctly, and units cancel as desired. The formula is reasonable because +5 is a common oxidation state of phosphorus. Practice 1. A compound is found to contain 63.52% iron and 36.48% sulfur. Problems Find its empirical formula. 2. Find the empirical formula of a compound found to contain 26.56% potassium, 35.41% chromium, and the remainder oxygen. 3. Analysis of 20.0 g of a compound containing only calcium and bromine indicates that 4.00 g of calcium are present. What is the empirical formula of the compound formed? Last objective, Objective 4: Determine a molecular formula from an empirical formula. Molecular Formula: the actual number and types of atoms in a molecule; may not be the same as the empirical formula. An empirical formula tells how many atoms of each element are combined in the simplest unit of a chemical compound. Sometimes the Molecular Formula is a multiple of the empirical formula; and you have to use collected data to calculate it. There is a ratio of the molecular formula to empirical formula, Let X be the ratio of the empirical formula mass to the molecular formula mass then:  (X) Empirical Formula = Molecular Formula For example: You have an empirical formula of P2O5 and molar mass of 283.89 g/mole; what is the molecular formula? Given: The empirical formula is P2O5; molecular formula mass is 283.89 g/mol Plan: Calculate empirical formula mass for P2O5; then use it to find the ratio, and finally apply the ratio to calculate the molecular formula Calculate: P2O5: (2)(30.97 g) + (5)(16.00 g) = 141.94g X = 2.0000 Which means that the molecular formula is 2.00 times the empirical formula So the molecular formula is (2)(P2O5) = P4O10 X = molecular formula mass empirical formula mass X = 283.89 g/mol 141.94 g/mol Determine the molecular formula for a compound with an empirical formula of NH2 and a formula mass of 32.06 amu. Given: The empirical formula is NH2; molecular formula mass is 32.06 amu Plan: Calculate empirical formula mass for NH2; then use it to find the ratio, and finally apply the ratio to calculate the molecular formula Calculate: NH2: (1)(14.01 amu) + (2)(1.01 amu) = 16.03 amu X = 2.000 Which means that the molecular formula is 2.0 times the empirical formula So the molecular formula is (2)(NH2) = N2H4 Your turn: 1. Determine the molecular formula of the compound with an empirical formula of CH and a formula mass of 78.110 amu. 2. A compound has a formula mass of 34.00 amu is found to consist of 0.44 g H and 6.92 g O. Find its molecular formula. X = 32.06 amu 16.03 amu