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Chapter 3: Problem 77

A compound that contains only carbon, hydrogen, and oxygen is \(48.64 \% \mathrm{C}\) and \(8.16 \% \mathrm{H}\) by mass. What is the empirical formula of this substance?

Short Answer

Expert verified
The empirical formula for the given compound, containing only carbon, hydrogen, and oxygen with 48.64% C and 8.16% H by mass, is C鈧僅鈧哋鈧�.

Step by step solution

01

Calculate the percentage of oxygen by mass

From the given information, we know the percentage of carbon and hydrogen. The compound contains only carbon, hydrogen, and oxygen, so we can calculate the percentage of oxygen by subtracting the percentages of carbon and hydrogen from 100%: Percentage of oxygen = 100% - (48.64% + 8.16%) = 100% - 56.8% = 43.2% Now, we have the following information: Carbon: 48.64% Hydrogen: 8.16% Oxygen: 43.2%
02

Convert percentages to grams

Assume that we have a 100 gram sample of the compound. Then the mass of each element in the sample is: Carbon: 48.64 g Hydrogen: 8.16 g Oxygen: 43.2 g
03

Convert grams to moles

To find the mole ratio, we first need to convert the mass of each element to moles. We can do this using the molar mass of each element: Moles of Carbon = \( \frac{48.64 g}{12.01g/mol}\) = 4.05 mol Moles of Hydrogen = \( \frac{8.16 g}{1.01g/mol}\)= 8.08 mol Moles of Oxygen = \( \frac{43.2 g}{16.00g/mol}\) = 2.7 mol
04

Find the simplest whole number ratio

To find the simplest whole number ratio between the moles of the elements, divide all values by the smallest number of moles: Divide all values by 2.7: Carbon: \( \frac{4.05}{2.7}\) 鈮� 1.5 Hydrogen: \( \frac{8.08}{2.7}\) 鈮� 3 Oxygen: \( \frac{2.7}{2.7}\) = 1 We see that the ratio for carbon is not a whole number (1.5). To convert it into a whole number, multiply all ratios by 2: Carbon: 1.5 脳 2 = 3 Hydrogen: 3 脳 2 = 6 Oxygen: 1 脳 2 = 2
05

Write the empirical formula

Now that we have the whole number ratios for each element, we can write the empirical formula for the compound: C鈧僅鈧哋鈧� The empirical formula for the given compound is C鈧僅鈧哋鈧�.

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Most popular questions from this chapter

What amount (moles) is represented by each of these samples? a. \(150.0 \mathrm{~g} \mathrm{Fe}_{2} \mathrm{O}_{3}\) b. \(10.0 \mathrm{mg} \mathrm{NO}_{2}\) c. \(1.5 \times 10^{16}\) molecules of \(\mathrm{BF}_{3}\)

Many homes in rural America are heated by propane gas, a compound that contains only carbon and hydrogen. Complete combustion of a sample of propane produced \(2.641 \mathrm{~g}\) of carbon dioxide and \(1.442 \mathrm{~g}\) of water as the only products. Find the empirical formula of propane.

ABS plastic is a tough, hard plastic used in applications requiring shock resistance. The polymer consists of three monomer units: acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{~N}\right)\), butadiene \(\left(\mathrm{C}_{4} \mathrm{H}_{6}\right)\), and styrene \(\left(\mathrm{C}_{8} \mathrm{H}_{8}\right)\). a. A sample of \(\mathrm{ABS}\) plastic contains \(8.80 \% \mathrm{~N}\) by mass. It took \(0.605 \mathrm{~g}\) of \(\mathrm{Br}_{2}\) to react completely with a \(1.20-\mathrm{g}\) sample of \(\mathrm{ABS}\) plastic. Bromine reacts \(1: 1\) (by moles) with the butadiene molecules in the polymer and nothing else. What is the percent by mass of acrylonitrile and butadiene in this polymer? b. What are the relative numbers of each of the monomer units in this polymer?

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ a. What is the maximum mass of ammonia that can be produced from a mixture of \(1.00 \times 10^{3} \mathrm{~g} \mathrm{~N}_{2}\) and \(5.00 \times 10^{2} \mathrm{~g} \mathrm{H}_{2} ?\) b. What mass of which starting material would remain unreacted?

Cumene is a compound containing only carbon and hydrogen that is used in the production of acetone and phenol in the chemical industry. Combustion of \(47.6 \mathrm{mg}\) cumene produces some \(\mathrm{CO}_{2}\) and \(42.8 \mathrm{mg}\) water. The molar mass of cumene is between 115 and \(125 \mathrm{~g} / \mathrm{mol}\). Determine the empirical and molecular formulas.
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