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Chapter 5: Problem 83

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

Short Answer

Expert verified
The empirical formula of the substance is \(C_1H_2O_1\), which can be simplified to \(CHO\).

Step by step solution

01

Calculate the percentage of oxygen atoms

We are given the percentage by mass for carbon and hydrogen. Since the compound contains only carbon, hydrogen, and oxygen, we can calculate the percentage of oxygen by subtracting the sum of carbon and hydrogen percentages from 100%: Percentage of oxygen = 100% - (48.64% C + 8.16% H) Now, plug in the given percentages and compute the percentage of oxygen.
02

Convert the percentages to moles

To find the ratio of moles of carbon (C), hydrogen (H), and oxygen (O), we must convert the percentages of each element into moles. We can do this by dividing the percentages by the respective atomic masses of the elements. - Atomic mass of C = 12.01 g/mol - Atomic mass of H = 1.008 g/mol - Atomic mass of O = 16.00 g/mol Moles of C = (48.64% / 12.01 g/mol) = x mol Moles of H = (8.16% / 1.008 g/mol) = y mol Moles of O = (z% / 16.00 g/mol) = w mol
03

Find the ratio of moles of each element

Now, we have to find the ratio x:y:w (moles of C: moles of H: moles of O). To do this, divide each mole value by the smallest mole value obtained in step 2. Then, round the ratios to the nearest whole number.
04

Write the empirical formula

Based on the ratio x:y:w determined in step 3, find the empirical formula of the compound by writing the elements with their whole-number ratios as subscripts in the formula. The final result will give the simplest formula (empirical formula) for the compound.

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Most popular questions from this chapter

When \(\mathrm{M}_{2} \mathrm{S}_{3}(s)\) is heated in air, it is converted to \(\mathrm{MO}_{2}(s) .\) A \(4.000-\mathrm{g}\) sample of \(\mathrm{M}_{2} \mathrm{S}_{3}(s)\) shows a decrease in mass of \(0.277 \mathrm{g}\) when it is heated in air. What is the average atomic mass of \(\mathrm{M} ?\)

Express the composition of each of the following compounds as the mass percents of its elements. a. formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\) b. glucose, \(C_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) c. acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) plus other impurities. \(\mathrm{A}\) \(752-\mathrm{g}\) sample of impure iron ore is heated with excess carbon, producing \(453 \mathrm{g}\) of pure iron by the following reaction: $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}(g)$$ What is the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample? Assume that \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is the only source of iron and that the reaction is \(100 \%\) efficient.

When the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). In a certain experiment, \(20.00 \mathrm{g}\) iron metal was reacted with \(11.20 \mathrm{g}\) oxygen gas. After the experiment, the iron was totally consumed, and 3.24 g oxygen gas remained. Calculate the amounts of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) formed in this experiment.

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?
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