A compound Y containg C, H and O only was burnt in a strem of pure oxygen.The carbon(iv)oxide and water produced were collected in pure weighed reading obtained are initial mass of Y=20.63mg mass of CO2 product=57.94mg mass of water=11.85mg.What is the emperical fomular of the compound Y? [H=1? C=12, O=16].?

you can't count the mass of the O in CO2 and H2O as coming frm the compound (It was burned in O2). but you can deduce it .

mass of C in 57.94 grams of CO2
massC=12/(32+12) *57.94= ....
then convert that to moles of C
mass H in 11.85g water: 2/(18)*11.85=.... then convert that to moles H
mass O in original y= 20.63-mass H - mass C
convert that to moles of O
now you have moles of each in the original Y.
find the ratio of them, that leads directly to the empiricalformula

To find the empirical formula of compound Y, we need to determine the ratio of the elements present in the compound.

First, let's calculate the moles of CO2 and water produced.

Moles of CO2 = mass of CO2 / molar mass of CO2
Molecular mass of CO2 = molar mass of C + 2 * molar mass of O
= 12 + 2 * 16
= 44 g/mol

Moles of CO2 = 57.94 mg / 44 g/mol
= 0.00132 mol

Moles of water = mass of water / molar mass of water
= 11.85 mg / (2 * 1 + 16)
= 0.658 mol

Next, let's calculate the moles of carbon and hydrogen based on the reaction:
C + O2 → CO2
2H2 + O2 → 2H2O

From the balanced equation, we can see that the ratio of carbon to CO2 is 1:1, and the ratio of hydrogen to water is 2:2.

Therefore, the moles of carbon in compound Y are equal to the moles of CO2 produced, which is 0.00132 mol.

The moles of hydrogen in compound Y are half the moles of water produced, which is 0.329 mol (0.658 mol / 2).

Now, let's calculate the moles of oxygen.

Moles of oxygen = Moles of CO2 * number of oxygen atoms in CO2 + Moles of water * number of oxygen atoms in water

Number of oxygen atoms in CO2 = 2
Number of oxygen atoms in water = 1

Moles of oxygen = 0.00132 mol * 2 + 0.658 mol * 1
= 0.660 mol

Now, we have the moles of carbon, hydrogen, and oxygen in compound Y. We can now determine the empirical formula.

The empirical formula consists of the simplest whole number ratio between the elements. To find this ratio, we divide the moles of each element by the smallest number of moles obtained.

Moles of carbon / 0.00132 = 1
Moles of hydrogen / 0.00132 = 0.329 / 0.00132 = 250
Moles of oxygen / 0.00132 = 0.660 / 0.00132 = 500

So, the empirical formula of compound Y is CH2O.