Questions: Empirical formula from % composition (Step 1) Given that a compound X has the following percent composition: 40.00 % C 6.71 % H 53.28 % O That means that in a 100.00 g sample of compound X, you'd have type your answer... g of C type your answer... g of H type your answer... g of O type numbers into the boxes, watch your sig-figs and round correctly Empirical formula from % composition (Step 2) Given that a compound X has the following percent composition: 40.00 % C 6.71 % H 53.28 % O That means that in a 100.00 g sample of compound X, you'd have type your answer... mol of C type your answer... mol of H type your answer... mol of O use average atomic masses from ptable.com type numbers into the boxes, watch your sig-figs and round correctly

Empirical formula from % composition (Step 1)
Given that a compound X has the following percent composition:
40.00 % C
6.71 % H
53.28 % O
That means that in a 100.00 g sample of compound X, you'd have
type your answer... g of C
type your answer... g of H
type your answer... g of O
type numbers into the boxes, watch your sig-figs and round correctly

Empirical formula from % composition (Step 2)
Given that a compound X has the following percent composition:
40.00 % C
6.71 % H
53.28 % O
That means that in a 100.00 g sample of compound X, you'd have
type your answer... mol of C
type your answer... mol of H
type your answer... mol of O
use average atomic masses from ptable.com
type numbers into the boxes, watch your sig-figs and round correctly

Transcript text: Empirical formula from \% composition (Step 1) Given that a compound X has the following percent composition: $40.00 \%$ C $6.71 \% \mathrm{H}$ $53.28 \%$ O That means that in a 100.00 g sample of compound X , you'd have type your answer... g of C type your answer... g of H type your answer... g of $O$ **type numbers into the boxes, watch your sig-figs and round correctly Empirical formula from \% composition (Step 2) Given that a compound X has the following percent composition: $40.00 \%$ C $6.71 \% \mathrm{H}$ $53.28 \%$ O That means that in a 100.00 g sample of compound X , you'd have type your answer... mol of $C$ type your answer... mol of H type your answer... mol of $O$ **use average atomic masses from ptable.com **type numbers into the boxes, watch your sig-figs and round correctly

Solution

Solution Steps

Step 1: Calculate Mass of Each Element in 100 g Sample

Given the percent composition of compound X, we can calculate the mass of each element in a 100 g sample:

Mass of Carbon (C): $40.00\% \times 100.00 \, \text{g} = 40.00 \, \text{g}$
Mass of Hydrogen (H): $6.71\% \times 100.00 \, \text{g} = 6.71 \, \text{g}$
Mass of Oxygen (O): $53.28\% \times 100.00 \, \text{g} = 53.28 \, \text{g}$

Step 2: Calculate Moles of Each Element

To find the moles of each element, we use their respective atomic masses:

Atomic mass of Carbon (C): $12.01 \, \text{g/mol}$
Atomic mass of Hydrogen (H): $1.008 \, \text{g/mol}$
Atomic mass of Oxygen (O): $16.00 \, \text{g/mol}$

Calculate the moles for each element:

Moles of Carbon (C): \[ \frac{40.00 \, \text{g}}{12.01 \, \text{g/mol}} = 3.3314 \, \text{mol} \]
Moles of Hydrogen (H): \[ \frac{6.71 \, \text{g}}{1.008 \, \text{g/mol}} = 6.6573 \, \text{mol} \]
Moles of Oxygen (O): \[ \frac{53.28 \, \text{g}}{16.00 \, \text{g/mol}} = 3.3300 \, \text{mol} \]