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Chapter 3: Problem 50

What are the empirical formulas of the compounds with the following compositions? (a) 40.1 percent \(\mathrm{C}\) 6.6 percent \(\mathrm{H}, 53.3\) percent \(\mathrm{O}\) (b) 18.4 percent \(\mathrm{C}\) 21.5 percent \(\mathrm{N}, 60.1\) percent \(\mathrm{K}\)

Short Answer

Expert verified
The empirical formula for the compound with 40.1 percent C, 6.6 percent H, and 53.3 percent O is \(\mathrm{CHO}_2\), while for the compound 18.4 percent C, 21.5 percent N, and 60.1 percent K, it is \(\mathrm{CNK}\).

Step by step solution

01

Conversion to grams

Assume you have a 100g sample of each compound. Therefore, the percentage of each element directly translates to the mass in grams. (a) For compound one: \(\mathrm{C} = 40.1g\), \(\mathrm{H} = 6.6g\), \(\mathrm{O}= 53.3g\). (b) For compound two: \(\mathrm{C} = 18.4g\), \(\mathrm{N} = 21.5g\), \(\mathrm{K} = 60.1g\).
02

Conversion to moles

Next, convert these masses to moles using the atomic masses of the elements (grams per mole): (a) For compound one: Moles of \(\mathrm{C} = 40.1g / 12.01g/mol = 3.34 mol\), moles of \(\mathrm{H} = 6.6g / 1.01g/mol = 6.54 mol\), moles of \(\mathrm{O} = 53.3g / 16.00g/mol = 3.33 mol\). (b) For compound two: Moles of \(\mathrm{C} = 18.4g / 12.01g/mol = 1.53 mol\), moles of \(\mathrm{N} = 21.5g / 14.01g/mol = 1.53 mol\), moles of \(\mathrm{K} = 60.1g / 39.10g/mol = 1.54mol\).
03

Finding the empirical formula

Divide each of the numbers of moles by the smallest number among them to determine the ratio. If necessary, multiply these numbers to get whole numbers. (a) For compound one: The ratio of \(\mathrm{C : H : O}\) is \(3.34 : 6.54 : 3.33\). Dividing by the least number, the ratio of \(\mathrm{C : H : O}\) is \(1 : 1.96 : 1\). After multiplying by 2, the ratio becomes \(2 : 4 : 2\) or \(1 : 2 : 1\). Therefore, the empirical formula is \(\mathrm{CHO}_2\). (b) For compound two: The ratio of \(\mathrm{C : N : K}\) is \(1.53 : 1.53 : 1.54\). Dividing by the smallest number, the ratio becomes approximately \(1 : 1 : 1\), hence the empirical formula is \(\mathrm{CNK}\).

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Most popular questions from this chapter

An iron bar weighed 664 g. After the bar had been standing in moist air for a month, exactly one-eighth of the iron turned to rust \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\). Calculate the final mass of the iron bar and rust.

The empirical formula of a compound is \(\mathrm{CH}\). If the molar mass of this compound is about \(78 \mathrm{~g}\), what is its molecular formula?

A mixture of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) is heated until all the water is lost. If \(5.020 \mathrm{~g}\) of the mixture gives \(2.988 \mathrm{~g}\) of the anhydrous salts, what is the percent by mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) in the mixture?

Cinnamic alcohol is used mainly in perfumery, particularly in soaps and cosmetics. Its molecular formula is \(\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}\) (a) Calculate the percent composition by mass of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) in cinnamic alcohol. (b) How many molecules of cinnamic alcohol are contained in a sample of mass \(0.469 \mathrm{~g} ?\)

Calculate the molecular mass or formula mass (in amu) of each of the following substances: (a) \(\mathrm{CH}_{4}\) (b) \(\mathrm{NO}_{2},(\mathrm{c}) \mathrm{SO}_{3},\) (d) \(\mathrm{C}_{6} \mathrm{H}_{6}\) (e) NaI, (f) \(\mathrm{K}_{2} \mathrm{SO}_{4}\) \((\mathrm{g}) \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)
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