Question

An organic compound is known to be composed of 46.60% carbon, 6.800% hydrogen, and 46.60% oxygen. What is its empirical formula?

Answer 1

`C_4H_7O_3`

Explanation:

Step 1:

For this type of problem, if the percentages add up to 100% (and it almost always is), then it safe to assume 100g sample.

By assuming that the sample is 100g, we can then assume that the corresponding percentages are also the weights.

Despite the fact that, in your case, the sum is only 99.8%, we'll just go on with the assumption of 100g sample shall we?

Thus, mass of C = 46.60 g; mass of H = 6.80g; mass of O = 46.60g

Step 2:

Since chemical formulas deal more with number of moles rather than weights, we have to convert these masses by multiplying them with their respective atomic masses.

mole C = 46.60 `cancel g` x `(1mol)/(12.0107 cancel "g") C` = 3.8799 mol C

mole H = 6.80 `cancel g` x `(1mol)/(1.00794 cancel "g") H` = 6.7464 mol H

mole O = 46.60 `cancel g` x `(1mol)/(15.9994 cancel "g") O` = 2.9126 mol O

Step 3:

Now that we have the number of moles of each element, we have to get their respective ratios by dividing everything with the smallest number of mole.

`C` = `(3.8799 cancel "mol")/(2.9126 cancel "mol")` = `1.3321`

`H` = `(6.7464 cancel "mol")/(2.9126 cancel "mol")` = `2.3163`

`O` = `(2.9126 cancel "mol")/(2.9126 cancel "mol")` = `1.000`

Step 4:

Round off the answers to the nearest whole number. For the above answers, you can multiply each number by a factor to get the lowest whole number multiple. In this case, the factor is 3. Thus,

`C` = 1.3321 x 3 = `3.99~~4`

`H` = 2.3163 x 3 = `6.95~~7`

`O` = 1.000 x 3 = `3`

Therefore, answer is `C_4H_7O_3`