What is the empirical formula of the unknown compound?

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What is the empirical formula of the unknown compound?

An unknown compound contains only C, H, and O. Combustion of 4.90 g of this compound produced 12.0 g of CO2 and 4.90 g of H2O.

CHO (insert subscripts as needed)

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Answer 1 · 2016-11-29T21:25:54

$C_{3} H_{8} O$ $"C"_3"H"_8"O"$

Explanation:

Since I want you to get some practice yourself, I will be showing you the method but to a different question. Hopefully, it will help you to solve this one.

What is the empirical formula for isopropyl alcohol (which contains only C, H and O) if the combustion of $0.255$ $0.255$ $g$ $"g"$ of isopropyl alcohol sample produces $0.561$ $0.561$ $g$ $"g"$ of ${CO}_{2}$ $"CO"_2$ and $0.306$ $0.306$ $g$ $"g"$ of $H_{2} O$ $"H"_2"O"$ ?

Before we start working out, we have to assume that we have one mole of all reagents and products.

If we write the empirical formula of the compound as $C_{x} H_{y} O_{z}$ $"C"_x"H"_y"O"_z$ , we can write its combustion reaction as:

$C_{x} H_{y} O_{z} + O_{2} \to {CO}_{2} + H_{2} O$ $"C"_x"H"_y"O"_z+"O"_2 rarr "CO"_2+"H"_2"O"$

$n ({CO}_{2}) = \frac{0.561}{44.0} = 0.0128$ $"n"("CO"_2)=0.561/44.0=0.0128$

One compound of ${CO}_{2}$ $"CO"_2$ is made up of one carbon atom and two oxygen atoms. So if we have $0.0128$ $0.0128$ moles of carbon dioxide then we also have $0.0128$ $0.0128$ moles of carbon in our sample (since we are assuming that we have one mole of each product). So now we can work out the mass of carbon in our product.

$m = 0.0128 \times 12 = 0.154$ $"m"=0.0128xx12=0.154$ $g$ $"g"$

$n (H_{2} O) = \frac{0.306}{18.0} = 0.017$ $"n"("H"_2"O")=0.306/18.0=0.017$

One compound of $H_{2} O$ $"H"_2"O"$ is made up of one oxygen atom and two hydrogen atoms. So if we have $0.017$ $0.017$ moles of water then we also have $2 (0.017)$ $2(0.017)$ , or $0.034$ $0.034$ , moles of hydrogen in our sample. So now we can work out the mass of hydrogen in our product.

$m = 0.034 \times 1 = 0.034$ $"m"=0.034xx1=0.034$ $g$ $"g"$

Adding these two values together will give us the mass of our compound that $C$ $"C"$ and $H$ $"H"$ make up.

$1.54 + 0.034 = 0.188$ $1.54+0.034=0.188$ $g$ $"g"$

So this means that the remaining mass must be oxygen.

$0.255 - 0.188 = 0.067$ $0.255 - 0.188 = 0.067$ $g$ $"g"$

$n (O_{2}) = \frac{0.067}{16} = 0.0042$ $"n"("O"_2)=0.067/16= 0.0042$

Now we have the molar ratio of each element in our compound:

$C$ $"C"$ : $H$ $"H"$ : $O$ $"O"$

$0.0128 : 0.017 : 0.0042$ $0.0128:0.017:0.0042$

We must divide by the smallest value ( $0.0042$ $0.0042$ ) to normalise the ratios.

$3 : 8 : 1$ $3:8:1$

So the empirical formula of the unknowns compound is $C_{3} H_{8} O$ $"C"_3"H"_8"O"$

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What is the empirical formula of the unknown compound?

An unknown compound contains only C, H, and O. Combustion of 4.90 g of this compound produced 12.0 g of CO2 and 4.90 g of H2O.

CHO (insert subscripts as needed)

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What is the empirical formula of the unknown compound?

An unknown compound contains only C, H, and O. Combustion of 4.90 g of this compound produced 12.0 g of CO2 and 4.90 g of H2O. CHO (insert subscripts as needed)

1 Answer

Explanation:

Related questions

Impact of this question

An unknown compound contains only C, H, and O. Combustion of 4.90 g of this compound produced 12.0 g of CO2 and 4.90 g of H2O.

CHO (insert subscripts as needed)