Question

You have a 5.00 g sample of an unknown compound. After analysis you find that its percent composition is 54.5% carbon, 9.10% hydrogen and 36.4% oxygen.

What is the empirical formula for this compound?

If the molar mass of this compound is 132 g/mol, what is its molecular formula?

Answer 1

5g of the compound is present.

C ? 5x54.5/100=2.725
H ? 5x9.10/100 =0.455
O ?36.4 x5/100 =1.82

2) Convert the masses to moles:

C ? 2.725/ 12 = 0.2270
H ? 0.455g / 1 = 0.455
O ? 1.82g / 16 = 0.1137

3) Divide by the lowest, seeking the smallest whole-number ratio:

C ? 0.2270/ 0.1137= 2
H ? 0.455/ 0.1137= 4
O ? 0.1137/ 0.1137= 1

4) Write the empirical formula:

C₂H₄O

5) Determine the formula weight:from atomic weights of carbon ,oxygen,hydrogen

12x2+1x4+16x1 = 44.

Assume 100 g of the compound is present. This changes the percents to grams:

C ? 54.5 g
H ? 9.10 g
O ?36.4 g

2) Convert the masses to moles:

C ? 54.50 g / 12 = 4.5
H ? 9.10g / 1 = 9.10
O ? 36.4 g / 16 = 2.275

3) Divide by the lowest, seeking the smallest whole-number ratio:

C ? 5.4 / 2.275 = 2
H ? 9.10/ 2.275 = 4
O ? 2.275 / 2.275 = 1

4) Write the empirical formula:

C₂H₄O

5) Determine the formula weight:from atomic weights of carbon ,oxygen,hydrogen

12x2+1x4+16x1 = 44

The emprical formula is C2H4O from percentages

n=molar mass/formula weight=132/44=3

(C2H4O )3=C6H12O3 is the molecular formula