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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 O 2 -2^ with H it looks like it should be reduced. H 2 O 2 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 BH 3 , but the molecular formula is B 2 H 6. 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: H 2 O 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:
- Find the number of moles of each element present.
- Determine the whole number mole ratio.
- 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.
- 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
- 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
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
Step 2: Divide each mole number by the smallest mole number – this gives the whole number ratio for the elements in the compound.
Step 3: Use the mole ratio for the subscripts of each element in the formula. The empirical formula is PbC 8 H 20
This is an organic compound that used to be added to gasoline, but the lead is too dangerous.
(38.4 g Pb)
1 mol Pb = 0.185 mol Pb 207.2 g Pb
(17.8 g C)
1 mol C = 1.48 mol C 12.01 g C
(3.74 g H )
1 mol H = 3.70 mol H 1.01 g H
0.185 mol = 1 0.185 mol 1.48 mol = 8 0.185 mol 3.70 mol = 20 0.185 mol
Determine the molecular formula for a compound with an empirical formula of NH 2 and a formula mass of 32.06 amu. Given: The empirical formula is NH 2 ; molecular formula mass is 32.06 amu
Plan: Calculate empirical formula mass for NH 2 ; then use it to find the ratio, and finally apply the ratio to calculate the molecular formula
Calculate: NH 2 : (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 )(NH 2 ) = N 2 H 4
Your turn:
- Determine the molecular formula of the compound with an empirical formula of CH and a formula mass of 78.110 amu.
- 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
