Question

A compound is 40.0% C, 6.70% H, and 53.3% O by mass. Assume that we have a 100.-g sample of this compound. Part A What are the subscripts in the empirical formula of this compound? Enter the subscripts for C, H, and O, respectively, separated by commas (e.g., 5,6,7).

Answer 1

1, 2, 1

You are calculating the empirical formula of this chemical compound, which is the question with moles, molar mass, and number of moles.

first you divide the mass of carbon by its molar mass(relative formula mass)because there is a formula about moles state: number of moles=mass/molar mass. So, 40/12 is about 3.3. Then, the RFM of H is 1, so the number of mole is 6.7/1=6.7. The RFM of O is 16, whihc means the molar mass od oxygen is 16g, so 53.3/16, which is about 3.3 too.

Next, you get the number of moles in order is; 3.3, 6.7, 3.3. Now we need to look at the ratio between these numbers. 3.3 and 3.3 has a common factor of 1, so the subscript of them are both 1. next use 6.7/3.3, and the greatest common factor is 2, so the answer is 1,2,1.

To solve this kind of  questions, there are many steps:

Know what you are calculating about, it's about the empirical formula, so you need to find out the number of moles of each elements.

1) Find the RFM of the element, because that is the molar mass(mass of 1 mole) of this element.

2) number of moles= mass/molar mass. use this formula to help you get the number of moles of each element in this compound

3) look at the relationship between the number of moles of each elements. find out the ratio between them.

4)that 's the answer of this question.

If i got anything wrong, please tell me :)

Answer 2

Answer : The subscripts for C, H and O in the empirical formula of this compound is 1, 2 and 1

Solution : Given,

Mass of C = 40.0 g

Mass of H = 6.70 g

Mass of O = 53.3 g

Molar mass of C = 12 g/mole

Molar mass of H = 1 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40.0g}{12g/mole}=3.33moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.70g}{1g/mole}=6.70moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.3g}{16g/mole}=3.33moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.33}{3.33}=1$

For H = $\frac{6.70}{3.33}=2.01\approx 2$

For O = $\frac{3.33}{3.33}=1$

The ratio of C : H : O = 1 : 2 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_1H_2O_1=CH_2O$

Therefore, the subscripts for C, H and O in the empirical formula of this compound is 1, 2 and 1