Question

A compound is found to contain 39.12 % carbon , 8.772 % hydrogen , and 52.11 % oxygen by mass.

To answer the question, enter the elements in the order presented above.

QUESTION 1:

The empirical formula for this compound is

______

QUESTION 2:
The molar mass for this compound is 92.11 g/mol.

The molecular formula for this compound is ______

.

Answer 1

1)

we have mass of each elements as:

C: 39.12 g

H: 8.772 g

O: 52.11 g

Divide by molar mass to get number of moles of each:

C: 39.12/12.01 = 3.2573

H: 8.772/1.008 = 8.7024

O: 52.11/16.0 = 3.2569

Divide by smallest:

C: 3.2573/3.2569 = 1

H: 8.7024/3.2569 = 8/3

O: 3.2569/3.2569 = 1

Multiply by 3 to get simplest whole number ratio:

C: 1*3 = 3

H: 8.3 * 3 = 8

O: 1*3 = 3

  

So empirical formula is:C3H8O3

Answer: C3H8O3

2)

Molar mass of C3H8O3,

MM = 3*MM(C) + 8*MM(H) + 3*MM(O)

= 3*12.01 + 8*1.008 + 3*16.0

= 92.094 g/mol

Now we have:

Molar mass = 92.11 g/mol

Empirical formula mass = 92.094 g/mol

Multiplying factor = molar mass / empirical formula mass

= 92.11/92.094

= 1

So molecular formula is:C3H8O3

Answer: C3H8O3