A compound has 11.21 % H and 88.79 % O. The molar mass of this compound is 34.02 g. Calculate the Empirical formula (E. F) and Molecular formula (M. F) of this compound.

All you really need in order to be able to answer this question is a periodic table.

You could actually solve this one by doing a quick calculation using the molar masses of the two elements, hydrogen and oxygen.

Take a look at the given percent composition. Notice that the ratio between the mass of hydrogen and the mass of oxygen in this compound is about 1:8. This means that you get eight times more grams of oxygen than you do of hydrogen in any sample of this compound.

Now look in the periodic table. The ratio between the molar mass of hydrogen, 1.00794 g/mol, and the molar mass of oxygen, 15.9994 g/mol, is about 1:16.

This means that you can match the 1:8 ratio given by the percent composition by doubling the amount of hydrogen you have from one mole to two moles.

You can thus conclude that the empirical formula for this compound, which uses the smallest whole number ratio that exists between the elements that make up a compound, is

H2O1⇒H2O

Now, here's how you can prove this. Pick a sample of this compound - since you're dealing with percentages, a 100.0-g sample will make the calculations easier.

This 100.0-gs sample will contain

11.2 g of hydrogen88.8 g of oxygen

Use the molar masses of the two elements to figure out how many moles of each you have here

11.2g⋅1 mole H1.00794g=11.11 moles H

88.8g⋅1 mole O15.9994g=5.550 moles O

To get the mole ratio that exists between the two elements in the compound, divide both values by the smallest one

For H: 11.11moles5.550moles=2.002≈2

For O: 5.550moles5.550moles=1

Since 2:1 is the smallest whole number ratio that can exist for these two elements, the empirical formula will once again be

H2O1⇒H2O

i believe it is 40–40–20%

explanation:

it's filtration, but this is an assumption so my apologies if i got this