Question

10 points!! Please give me work with it and I will mark as brainlist.

Answer the question below. Use the rubric in the materials for help if needed.
A compound is 40.0 % C, 53.3 % 0, and 6.66 % H What is its molecular formula if its molecular mass is 60 g? Show all your work.

Answer 1

Answer: The molecular formula will be $C_2H_4O_2$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 40.0 g

Mass of O = 53.3 g

Mass of H = 6.66 g

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

Moles of H =$\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.66g}{1g/mole}=6.66moles$

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

For H = $\frac{6.66}{3.33}=2$

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

Hence the empirical formula is $COH_2$

The empirical weight of $COH_2$ = 1(12)+1(16)+2(1)= 30g.

The molecular weight = 60 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{60}{30}=2$

The molecular formula will be=$2\times CH_2O=C_2H_4O_2$