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Chapter 3: Problem 86

A sample of urea contains \(1.121 \mathrm{~g} \mathrm{~N}, 0.161 \mathrm{~g} \mathrm{H}, 0.480 \mathrm{~g} \mathrm{C}\), and \(0.640 \mathrm{~g} \mathrm{O}\). What is the empirical formula of urea?

Short Answer

Expert verified
The empirical formula of urea is NH_2CO.

Step by step solution

01

Determine the moles of each element in the sample

First, we need to calculate the number of moles of each element present in the sample. We can do this by dividing the mass of the element by its molar mass. The molar mass of N = 14.01 g/mol, H = 1.01 g/mol, C = 12.01 g/mol, and O = 16.00 g/mol. - For nitrogen (N): \(\frac{1.121 g}{14.01 g/mol} = 0.0800\) moles - For hydrogen (H): \(\frac{0.161 g}{1.01 g/mol} = 0.159\) moles - For carbon (C): \(\frac{0.480 g}{12.01 g/mol} = 0.0400\) moles - For oxygen (O): \(\frac{0.640 g}{16.00 g/mol} = 0.0400\) moles
02

Find the smallest mole value amongst the elements

Now, we need to find the smallest mole value amongst the elements: The smallest mole value is 0.0400 (C and O).
03

Determine the mole ratio of each element

Now, we need to find the ratio between the actual moles of each element and the smallest mole value. This will give us the mole ratio for each element: - For nitrogen (N): \(\frac{0.0800}{0.0400} = 2\) - For hydrogen (H): \(\frac{0.159}{0.0400} \approx 4\) - For carbon (C): \(\frac{0.0400}{0.0400} = 1\) - For oxygen (O): \(\frac{0.0400}{0.0400} = 1\)
04

Write the empirical formula

Finally, write the empirical formula using the mole ratios as subscripts: The empirical formula is NH_2CO.

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Most popular questions from this chapter

A sample of urea contains \(1.121 \mathrm{~g} \mathrm{~N}, 0.161 \mathrm{~g} \mathrm{H}, 0.480 \mathrm{~g} \mathrm{C}\), and \(0.640 \mathrm{~g} \mathrm{O}\). What is the empirical formula of urea?

Silicon is produced for the chemical and electronics industries by the following reactions. Give the balanced equation for each reaction. a. \(\mathrm{SiO}_{2}(s)+\mathrm{C}(s) \frac{\text { Electic }}{\text { arc furmae }} \mathrm{Si}(s)+\mathrm{CO}(g)\) b. Liquid silicon tetrachloride is reacted with very pure solid magnesium, producing solid silicon and solid magnesium chloride. c. \(\mathrm{Na}_{2} \mathrm{SiF}_{6}(s)+\mathrm{Na}(s) \rightarrow \operatorname{Si}(s)+\operatorname{NaF}(s)\)

A compound contains only \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{N}\). Combustion of \(35.0 \mathrm{mg}\) of the compound produces \(33.5 \mathrm{mg} \mathrm{CO}_{2}\) and \(41.1 \mathrm{mg}\) \(\mathrm{H}_{2} \mathrm{O}\). What is the empirical formula of the compound?

Glass is a mixture of several compounds, but a major constituent of most glass is calcium silicate, \(\mathrm{CaSiO}_{3-}\) Glass can be etched by treatment with hydrofluoric acid; HF attacks the calcium silicate of the glass, producing gaseous and water-soluble products (which can be removed by washing the glass). For example, the volumetric glassware in chemistry laboratories is often graduated by using this process. Balance the following equation for the reaction of hydrofluoric acid with calcium silicate. $$ \mathrm{CaSiO}_{3}(s)+\mathrm{HF}(a q) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$

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 \(3.24 \mathrm{~g}\) oxygen gas remained. Calculate the amounts of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) formed in this experiment.
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