Question

a compound contains 27.3% Carbon and 72.7% Oxygen by mass. What is the empirical formula

Answer 1

Step 1) 27.3% C becomes 27.3g of C, while 72.7% of O becomes 72.7g of O
2) Using the mole mass and Mr equation (moles=mass/Mr), convert each to moles. i.e. 27.3/12=2.275 mol of C. 72.7/16=4.54375 mol of O.
3) Divide each mole amount by the smallest, which is 2.275/2.275=1 (for C) and 4.54375/2.275=1.99 (for O)
4) This shows that it is a 1:2 ration, as 1.99 is essentially double 1.
Therefore the answer is CO2