Question

Question #ddd3d

Answer 1

The simplest (or "empirical") formula is `C_3H_8O_3`

Explanation:

First, we take the masses of the two products and convert these to moles:

moles `CO_2 = 1.448g-:44.0 g/"mol" = 0.0329` mol

Since `CO_2` molecules contain one C atom each, this means there is `0.0329` mol of C atoms in the products. Therefore, the glycerol also contained `0.0329` mol of carbon.

moles `H_2O= 0.790g-:18g/"mol"=0.0439` mol

Since `h_2O` molecules each contain ywo atoms of H, the product must contain `2xx0.0439 = 0.0878` mol of H atoms. Therefore, the glycerol also contained `0.0878` mol of H.

To find the amount of oxygen in the glycerol, we must first determine the total mass of the above amounts of C and H:

Mass of C = `0.0329 "mol" xx 12g/"mol" = 0.395g`

Mass of H = `0.0878 "mol"xx1.0 g/mol=0.0878g`

Total mass of C + H = `0.483g`

But the original sample was `1.010g`, so the mass we have not yet accounted for must be the oxygen atoms in glycerol.

`1.010 - 0.483 = 0.527g`

moles of O = `0.527g-:16g/"mol" = 0.0329` mol

Looking at the three quantities in ratio form, we have:

Moles `C:H:O = 0.0329:0.0878:0.0329`

To simplify this ratio, we divide each value by the smallest (0.0329). This gives us:

Moles `C:H:O = 1:2.67:1`

Now, multiply each term by 3 (because the ".67" part of the H term is two-thirds) to get whole numbers:

Moles `C:H:O = 3:8:3`

and the simplest (or "empirical") formula is `C_3H_8O_3`