We can't tell. What's the other 90 percent? If you meant 40/60 instead... the mass of sulfur is twice that of oxygen, so a mass ratio of 40:60 is equivalent to an atom ratio of 1:3. The empirical formula would be SO3.
As2O3
BaCl2
Well first we assume a 100 g sample and then convert the elements into moles. We then divide them all by the number of moles that is the lowest of the 3. This procedure gives us AlPO4
We assume 100 grams and turn those percentages into grams and find moles of species. 40 grams carbon (1 mole C/12.01 grams) = 3.33 moles C 6.72 grams hydrogen (1 mole H/1.008 grams) = 6.67 moles H 53.28 grams oxygen (1 mole O/16 grams) = 3.33 The smallest mole number is 1, so we have two small numbers. Divide the large number by the small. 6.67/3/33 = 2 so.............. CH2O is the empirical formula. You did not state that you had a quantity of this compound ( such as 60 grams, or whatever ), so the molecular formula can not be found from this info. C6H12O6 is of course the molecular formula. To find the molecular formula you need a mass of the compound aside from the simple percentages. Then you find, as we have done, the empirical formula. you divide the mass given by the mass of the empirical formula and then take that quotient and multiply it times the empirical formula. In this case (CH2O) * 6 = C6H12O6, the molecular formula
The empirical formula of this compound would be MgO.
You should solve for an empirical formula when you are given the percent composition of elements in a compound or when you have the molar mass of the compound but not the molecular formula. The empirical formula provides the simplest whole-number ratio of atoms in a compound.
Chi a+
The percent composition of a compound with the empirical formula CO2 is 27.3% carbon and 72.7% oxygen.
CHI3
The empirical formula of the compound with 40% sulfur and 60% oxygen is SO3 (sulfur trioxide). The ratio of sulfur to oxygen in the compound is 1:3, which simplifies to SO3.
The empirical formula for the compound that is 43.6% carbon and 56.4% oxygen is CO2. This is because the ratio of carbon to oxygen in carbon dioxide is 1:2.
C4h2s
The empirical formula of the compound with 52.7% K and 47.3% Cl is KCl (potassium chloride). This is because the ratio of potassium to chlorine atoms in the compound is 1:1, leading to the simple formula KCl.
The empirical formula for a compound containing 13% magnesium and 87% bromine is MgBr2. This is because the ratio of magnesium to bromine atoms in the compound is 1:2, which corresponds to the formula MgBr2.
The empirical formula for this compound is CaSO4. This can be determined by converting the percentages to moles, finding the mole ratio, and simplifying the subscripts to the smallest whole number ratio.
Percent composition can be used to calculate the percentage of an element/compound in a mixture. From the percent composition, you can also find the empirical formula. And from the empirical formula you can find the actual molecular weight.