Burning 12.00 g of an oxoacid produces 17.95 g of carbon dioxide and 4.87 g of water. Consider that 0.25 moles of oxoacid equals 44.0 g. For this compound, determine the empirical and molecular formula.

The molecular formula will be

Explanation:

Mass of = 17.95 g

Mass of = 4.87 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 17.95 g of carbon dioxide, = of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.87 g of water, = of hydrogen will be contained.

Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g

Mass of C = 4.89 g

Mass of H =  0.541 g

Mass of O = 6.57 g

Step 1 : convert given masses into moles.

Moles of C =

Moles of H=

Moles of O=

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =

For H =

For O =

The ratio of C : H : O = 1: 1  : 1

Hence the empirical formula is .

Hence the empirical formula is

The empirical weight of = 1(12)+1(1)+1(16)= 29 g.

If 0.25 moles has mass of 44.0 g

Thus 1 mole has mass of =

Thus molecular mass is 176 g

Now we have to calculate the molecular formula.

The molecular formula will be=