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Chapter 9: Problem 88

An organic compound of carbon and hydrogen has the empirical formula CH. What is its molecular formula if its molar mass is: (a) \(26 \mathrm{~g} / \mathrm{mol}\) (b) \(52 \mathrm{~g} / \mathrm{mol}\) (c) \(78 \mathrm{~g} / \mathrm{mol}\)

Short Answer

Expert verified
The molecular formulas for the given molar masses are: (a) \(C_{2}H_{2}\) for 26 g/mol (b) \(C_{4}H_{4}\) for 52 g/mol (c) \(C_{6}H_{6}\) for 78 g/mol

Step by step solution

01

Find the empirical formula's molar mass

The given empirical formula is CH, which contains one carbon atom and one hydrogen atom. The molar mass of carbon (C) is 12 g/mol, and the molar mass of hydrogen (H) is 1 g/mol. Therefore, the molar mass of the empirical formula (CH) is: \( Molar~mass~of~CH = 12 + 1 = 13 ~g/mol \)
02

Determine the whole number multiplier

Taking the given molar masses of the molecular formula and dividing each by the empirical formula's molar mass will provide the whole number multiplier we'll require in the next step. (a) \( Whole~number~multiplier = \frac{26 ~g/mol}{13 ~g/mol} = 2 \) (b) \( Whole~number~multiplier = \frac{52 ~g/mol}{13 ~g/mol} = 4 \) (c) \( Whole~number~multiplier = \frac{78 ~g/mol}{13 ~g/mol} = 6 \)
03

Find the molecular formula using the whole number multiplier

Now, we can multiply the empirical formula (CH) by the whole number multiplier to find the molecular formula. (a) Molecular Formula: \(2 \times (CH) = \) \(C_{2}H_{2}\) (b) Molecular Formula: \(4 \times (CH) = \) \(C_{4}H_{4}\) (c) Molecular Formula: \(6 \times (CH) = \) \(C_{6}H_{6}\) So, the molecular formulas for the given molar masses are: (a) \(C_{2}H_{2}\) for 26 g/mol (b) \(C_{4}H_{4}\) for 52 g/mol (c) \(C_{6}H_{6}\) for 78 g/mol

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Most popular questions from this chapter

Determine the empirical formula of the compound with the following mass percents of the elements present: \(58.5 \% \mathrm{C} ; 4.91 \% \mathrm{H} ; 19.5 \% \mathrm{O} ; 17.1 \% \mathrm{~N}\).

The compound ibuprofen, used in some pain relievers, has the molecular formula \(\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2}\). If \(0.250 \mathrm{~g}\) of ibuprofen is burned in oxygen, how many grams of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) will be produced?

Acetaminophen, used in many over-the-counter pain relievers, has a molecular formula of \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{NO}_{2} .\) Calculate the mass percentage of each element in acetaminophen.

Butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\), used as the fuel in disposable lighters, reacts with oxygen \(\left(\mathrm{O}_{2}\right)\) to produce \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). Suppose \(10.00 \mathrm{~g}\) butane is combined with \(10.00 \mathrm{~g} \mathrm{O}_{2}\). (a) Write a balanced chemical equation for the combustion of butane. (b) Which reactant is limiting? (c) What is the theoretical yield of each product for this reaction in grams? (d) How many grams of excess reactant are left over at the end of the reaction? (e) How many additional grams of the limiting reactant are required to run this reaction in a balanced fashion?

Nitrogen and fluorine react to form nitrogen trifluoride according to the balanced chemical equation \(\mathrm{N}_{2}(g)+3 \mathrm{~F}_{2}(g) \rightarrow 2 \mathrm{NF}_{3}(g)\) For each of the following reaction mixtures, choose the limiting reactant: (a) \(0.50 \mathrm{~mol} \mathrm{~N}_{2}(g)\) and \(0.50 \mathrm{~mol} \mathrm{~F}_{2}(g)\) (b) \(12.0 \mathrm{~mol} \mathrm{~N}_{2}(g)\) and \(20.0 \mathrm{~mol} \mathrm{~F}_{2}(g)\) (c) \(2.5 \mathrm{~mol} \mathrm{~N}_{2}(g)\) and \(7.5 \mathrm{~mol} \mathrm{~F}_{2}(g)\) (d) 100 molecules \(\mathrm{N}_{2}(g)\) and 500 molecules \(\mathrm{F}_{2}(g)\) (e) \(5.00 \mathrm{~g} \mathrm{~N}_{2}(g)\) and \(15.0 \mathrm{~g} \mathrm{~F}_{2}(g)\) (f) \(20.0 \mathrm{mg} \mathrm{N}_{2}(g)\) and \(70.0 \mathrm{mg} \mathrm{F}_{2}(g)\)
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