A 11.0g sample of compound containing only oxygen and carbon produces 3.0g carbon and 8.0g oxygen. Determine the empirical formula of the compound.
3g C = 1/4 mole
8g O2 = 1/2 mole
So, you have CO2
CxOy ==> xC + 1/2O2
% C = (3/11)*100 = ? about 27
% O2 = (8/11(*100 = about 73
mols C = about 27/12 = about 2.25
mols O2 = about 73/32 = about 2.28 O2 molecules or about 4.56 mols O atoms.
Now find the ratio of the C atoms to the O ATOMS, round to whole numbers and that will be the empirical formula. The easy way to find the ratio is to divide the smaller number by itself (which makes it 1.00), then divide the other number by the same small number.
show the all working out of the particular question .
To determine the empirical formula of a compound, we need to find the ratio of the elements present in the compound. In this case, we have the masses of carbon and oxygen, so we can use them to find the ratio.
Step 1: Calculate the moles of carbon and oxygen.
To calculate the moles, we can use the formula:
moles = mass / molar mass
- Carbon:
The molar mass of carbon (C) is 12.01 g/mol. Using this molar mass, we can calculate the moles of carbon:
moles of carbon = 3.0g / 12.01 g/mol
- Oxygen:
The molar mass of oxygen (O) is 16.00 g/mol. Using this molar mass, we can calculate the moles of oxygen:
moles of oxygen = 8.0g / 16.00 g/mol
Step 2: Find the simplest whole-number ratio.
To find the simplest whole-number ratio, we need to divide the moles of each element by the smallest number of moles.
In this case, the moles of carbon is 0.2495 and the moles of oxygen is 0.5.
Dividing both moles by 0.2495 (the smallest number of moles), we get the following ratio:
moles of carbon / 0.2495 = 0.2495 / 0.2495 = 1
moles of oxygen / 0.2495 = 0.5 / 0.2495 ≈ 2
Therefore, the empirical formula of the compound is C1O2, or simply CO2.