Determining Chemical Formulas
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1 SECTION 7.4 Determining Chemical Formulas When scientists discover or produce a new compound, they analyze it to learn its composition. Often what they measure is the percentage composition of the substance. Using the molar masses of the elements in the compound, they can convert the percentages into mole ratios. These molar ratios give the empirical formula for a compound. An empirical formula consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole-number mole ratio of the different elements. An empirical formula is often the same as the compound s chemical formula. However, some molecular compounds, such as diborane, B 2 H 6, consist of molecules that do not have the lowest whole-number ratio of atoms possible. The empirical formula for diborane is BH 3. empirical formula Key Terms For ionic compounds, the empirical formula is almost always the same as the formula unit. However, the empirical formula would not reflect the presence of a polyatomic ion. For example, the empirical formula of Ba(NO 3 ) 2 is BaN 2 O 6. TIP Empirical formulas show the whole number ratio of elements in a compound. Empirical formulas are calculated from the percentage composition of a sample. The following steps will convert a percentage composition into an empirical formula. 1. Assume that you have 100 g of the sample. The mass of each element in this sample is numerically equivalent to the percentage composition for that element. 2. Convert each mass into a number of moles by using each element s molar mass as a conversion factor. Express the result as a molar ratio of the elements in the sample. 3. Divide the ratio from Step 2 by the number of moles of the element with the least amount of moles. If the result is a ratio of numbers very close to whole numbers, then this ratio gives the empirical formula of the sample. 4. If Step 3 does not yield a ratio of whole numbers, multiply the ratio by the smallest number that will give a whole-number ratio. Critical Thinking 1. Apply An analysis of a sample of a compound determines that the compound contains 1 mol of phosphorus atoms for every mol of oxygen atoms. What is the empirical formula of the compound? Chemical Formulas and Chemical Compounds 235
2 SAMPLE PROBLEM Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula of the compound. SOLUTION 1 ANALYZE Determine the information that is given and unknown. Given: Sample is 32.38% Na, 22.65% S, 44.99% O by mass. Unknown: empirical formula 2 PLAN Describe the steps that will be used to find the unknown. The procedure on the previous page will convert a percentage composition into an empirical formula. 3 SOLVE Calculate the unknown using the given information. A g sample of the compound contains g Na, g S, and g O g Na = g Na 1 mol Na = mol Na g Na g S = g S mol g S = mol S S g O = g O mol g O = mol O O Then, divide the mole ratio of the three elements by , which is the least of the three amounts above mol Na : mol S : mol O = mol Na : mol S : mol O This ratio is close to the ratio of whole numbers 2:1:4. Therefore, the empirical formula of the compound is Na 2 SO 4. 4 CHECK YOUR WORK Check the answer to see if it is reasonable. The result is an ionic compound formed from Na + cations and SO anions, which is a plausible combination. 236 CHAPTER 7
3 PRACTICE A. A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula. What mass of Fe is present in a 100 g sample of this compound? What mass of S is present in a 100 g sample of this compound? Convert each mass above into moles. g Fe = g Fe 1 mol Fe = g Fe g S = g S 1 mol S = g S Write the results above as a ratio and divide by the smallest amount. mol Fe: mol S, divided by = mol Fe: mol S What is the empirical formula of the compound? B. Find the empirical formula of a compound containing 26.56% potassium, 35.41% chromium, and the remainder oxygen. What mass of K is present in a 100 g sample of this compound? What mass of Cr is present in a 100 g sample of this compound? What mass of O is present in a 100 g sample of this compound? Convert each mass above into moles. g K = g K 1 mol K = g K g Cr = g Cr 1 mol Cr = g Cr g O = g O 1 mol O = g O Write the results above as a ratio and divide by the smallest amount. divided by mol K: mol Cr: mol O, = mol K: mol Cr: mol O What is the empirical formula of the compound? Chemical Formulas and Chemical Compounds 237
4 Molecular formulas give the types and numbers of atoms in a compound. The empirical formula may or may not be equivalent to the molecular formula of a compound. Some compounds that have different molecular formulas have the same empirical formula. For example, examine the formulas for the two compounds, ethene and cyclopropane. Ethene: molecular formula: C 2 H 4 empirical formula: CH 2 Cyclopropane: molecular formula: C 3 H 6 empirical formula: CH 2 The empirical formula is always the smallest whole-number mole ratio of the elements in a compound. A molecular formula does not have to be the smallest ratio, but it must be a whole-number multiple of the smallest ratio. In other words, x(empirical formula) = molecular formula where x is some whole number greater than or equal to 1. The numbers of atoms of each element in the empirical formula are each multiplied by x to obtain the molecular formula. In order to determine the value of x, a chemist must have an additional piece of information. The chemist must be able to calculate the formula mass of the compound. The relationship between the empirical formula mass and the molecular formula mass is given by! Remember The formula mass of a compound in u is equal to its molar mass in g/mol. x(empirical formula mass) = molecular formula mass where x is the same whole number greater than or equal to 1 as defined in the first equation. For example, from percentage composition measurements, a chemist might determine that the empirical formula of a compound is BH 3, with an empirical formula mass of u. If separate measurements determine that the formula mass of the compound is u, then the value of x is x = molecular formula mass empirical formula mass = u = u Therefore, the chemical formula of the compound is B 2 H 6, and the compound is diborane. READING CHECK 2. Calculate The empirical formula for ethene, C 2 H 4, and cyclopropane, C 3 H 6, is CH 2. Use the relationship given above to calculate the value of x for each compound. ethene: cyclopropane: 238 CHAPTER 7
5 SAMPLE PROBLEM The empirical formula of a compound of phosphorus and oxygen is P 2 O 5. Experimentation shows that the molar mass of this compound is g/mol. What is the compound s molecular formula? SOLUTION 1 ANALYZE Determine the information that is given and unknown. Given: empirical formula Unknown: molecular formula 2 PLAN Give the equations that will be used to find the unknown. First, the value of x can be determined by this relationship. x = molecular formula mass empirical formula mass Then, x can be used to calculate the molecular formula. x(empirical formula) = molecular formula 3 SOLVE Calculate the unknown using the given information. The molar mass of the compound is g/mol, so the molecular formula mass is u. Mass of P in P 2 O 5 = 2 P atoms P atom u = u Mass of O in P 2 O 5 = 5 O atoms O atom u = u empirical formula mass = u x = molecular formula mass empirical formula mass = u u = Therefore, the molecular formula is 2(P 2 O 5 ) = P 4 O CHECK YOUR WORK Check the answer to see if it is reasonable. The empirical formula mass is about equal to = 140, which is half of the given molar mass. Chemical Formulas and Chemical Compounds 239
6 PRACTICE C. Determine the molecular formula of the compound with an empirical formula of CH and a formula mass of u. The empirical formula mass of CH is u + u = u x = molecular formula mass empirical formula mass = u = The molecular formula is (CH) =. D. A sample of a compound with a formula mass of u is found to consist of 0.44 g H and 6.92 g O. Find its molecular formula. g H = g H 1 mol H g H = g O = g O 1 mol O g O = Write the results above as a ratio and divide by the smallest amount. mol H: mol O, divided by = mol H: mol O What is the empirical formula of the sample? The empirical formula mass of the sample is u + u = u x = molecular formula mass empirical formula mass = u = The molecular formula is =. 240 CHAPTER 7
7 SECTION 7.4 REVIEW VOCABULARY 1. What is the relationship between the empirical formula and the molecular formula of a compound? REVIEW 2. A compound contains 36.48% Na, 25.41% S, and 38.11% O. Find its empirical formula. 3. If 4.04 g of N combine with g O to produce a compound with a formula mass of u, what is the molecular formula of the compound? Critical Thinking 4. RELATING IDEAS A compound containing sodium, chlorine, and oxygen is 25.42% sodium by mass. A 3.25 g sample gives atoms of oxygen. What is its empirical formula? Chemical Formulas and Chemical Compounds 241
8 Math Tutor Calculating Percentage Composition Chemists can analyze an unknown substance by determining its percentage composition by mass. Percentage composition is determined by finding the mass of each element in a sample as a percentage of the whole sample. The results of this analysis can then be compared with the percentage composition of known compounds to determine the probable identity of the unknown substance. Once you know a compound s formula, you can determine its exact percentage composition by mass. Problem-Solving TIPS To determine the percentage composition for a compound using its chemical formula, first determine the contribution of each element to the molar mass of the compound. Use this information to determine the molar mass of the compound. Then calculate the ratio of the mass contribution of each element to the total molar mass and express the ratio as a percentage. SAMPLE Determine the percentage composition of potassium chlorate, KClO 3. First, calculate the molar mass of KClO 3. Mass of 1 mol K = g K Mass of 1 mol Cl = g Cl Mass of 3 mol O = 3(16.00 g O) = g O The molar mass of KClO 3 is g g g = g. The percentage composition of KClO 3 is determined by using the following formula to calculate the percentage of each element. mass of element in 1 mol 100% = % element in compound molar mass Percent K by mass in KClO 3 = g K 100% = 31.91% g KClO 3 Percent Cl by mass in KClO 3 = g Cl 100% = 28.93% g KClO 3 Percent O by mass in KClO 3 = g O 100% = 39.17% g KClO 3 Practice Problems: Chapter Review practice problems 17 and CHAPTER 7
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