Help me out I need help in chemistry?

Question

If you are given a sample which you experimentally determine to be 40.000 % C, 6.714 % H, and 53.289 % O by weight, what is the empirical formula for the sample ?

Answer 1

That is correct you convert this percentages straight to masses (same # they are now) in a 100.0g sample. Then convert these masses to moles based on each atoms' molar mass. Divide each molar mass by the smaller one to get the numbers of each atom in the empirical formula. e.g. : C2H4O2

Answer 2

If I understand your question correctly then this should be the answer:

Assume 100g

Element/ Mass/ Molar Mass /    Moles    /   Ratio /   Whole # Ratio

C /40 g /12.01 g/mol     /3.33 mol      /1          /1

H /6.714 g /1.01 g/mol       /6.65 mol    / 1.99         / 2

O /53.289 g /16.00 g/mol     /3.33 mol        /1          /1

Empirical Formula: CH2O

Supposed to be a chart (seperated by /)

Answer 3

We consider 100 g of this compound

Moles C = 40.000 g / 12.011 =3.330

Moles H = 6.714 g / 1.008 =6.661

Moles O = 53.289 g / 15.9994 = 3.330

we get

C ( 3.330) H (6.661 ) O ( 3.330)

we divide by the smallest number

3.330 / 3.330 = 1 => C

6.661 / 3.330 = 2 => H

3.330 / 3.330 = 1 => O

the empirical formula is

CH2O