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: -2 example peroxide O2 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. 1 mol Na We cannot use (36.5g Na) = 1.59 mol Na 22.99 g Na decimal values 1 mol S for subscripts, (25.4g S) = 0.79 mol S 32.07 g S so on to step 2 1 mol O (38.1g O) = 2.38 mol O 16.00 g O
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 S = .79/.79 = 1 O = 2.38/.79 = 3 (2.012) (3.012)
3. Use the mole ratio for the subscripts of each element in the formula. The empirical formula is Na2SO3 Sodium Sulfite
Now it is your turn: Practice #1: If a compound has 32.38% sodium, 22.65% sulfur and 44.99% oxygen what would the empirical formula be? Step 1: Find the number of moles of each element present Part A 32.38% Na 32.38 g Na 22.65% S 22.65 g S 44.99% O 44.99 g O Part B: 1 mol Na (32.38 g Na) = 1.408 mol Na 22.99 g Na 1 mol S (22.65 g S) = 0.7063 mol S 32.07 g S 1 mol O (44.99 g O ) = 2.812 mol O 16.00 g O
Step 2: Divide each mole number 1.408 mol = 2 (1.993) by the smallest mole 0.7063 mol number – this gives the 0.7063 mol = 1 whole number ratio for the 0.7063 mol elements in the compound. 2.812 mol = 4 (3.981) 0.7063 mol Step 3: Use the mole ratio for the subscripts of each element in the formula. The empirical formula is Na2SO4 Sodium Sulfate Practice #2: A 60.0g sample of tetraethyl-lead contains 38.4 g lead, 17.8 g carbon, and 3.74 g hydrogen. Find its empirical formula. Step 1: Find the number of moles of each element present Part A already done. Part B 1 mol Pb (38.4 g Pb) = 0.185 mol Pb 207.2 g Pb 1 mol C (17.8 g C) = 1.48 mol C 12.01 g C 1 mol H (3.74 g H ) = 3.70 mol H 1.01 g H
Step 2: Divide each mole number 0.185 mol = 1 by the smallest mole 0.185 mol number – this gives the 1.48 mol = 8 whole number ratio for the 0.185 mol elements in the compound. 3.70 mol = 20 0.185 mol
Step 3: Use the mole ratio for the subscripts of each element in the formula. The empirical formula is PbC8H20
This is an organic compound that used to be added to gasoline, but the lead is too dangerous.
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: molecular formula mass X = empirical formula mass (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
283.89 g/mol X = X = 2.0000 Which means that the 141.94 g/mol molecular formula is 2.00 times the empirical formula
So the molecular formula is (2)(P2O5) = P4O10
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 32.06 amu X = X = 2.000 Which means that the molecular 16.03 amu 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.