A gaseous binary (2 distinct elements) compound which is known to contain hydrogen (H) has a density of 1.94 times that of oxygen gas (at the same temperature and pressure). When 1.31 g of the compund is completly burned in ecxess oxygen (O2), 1.21 g of water is formed. Determine the formula of the compound.

 

Ok...I'm pretty sure this is a combustion reaction, and that you are supposed to solve for moles using density. But how? This one is hard, I think, because it is difficult to see the relationships between the various elements.

Remember that the density of a gas is directly related to it's molecular weight. A high molecular weight gas will be much more dense than a low molecular weight gas. So if you assume your gas to be at STP, you can calculate the molar mass of it.

 

Since one mole of oxygen at STP is 32 grams divided by 22.4 liters which equals 1.43x10^-3 g/mL, you know that the density of your unknown gas at that temperature equals 2.77x10^-3 g/mL. If you have one mole, which would take up 22400 mL, you can figure out the molecular mass which is 62.08 grams per mol. You know the gas is binary, so it has a molecular formula of X(y)H(z). The question also tells you that 1.31 grams burned produces 1.21 grams of water. Hydrogen makes up about 11.111% of the mass of water, and we know that the hydrogen had to come from your compound. So if 1.21 grams of water was formed, then 0.1344 grams of hydrogen is in there. (Since 1.21*0.11111 equals 0.1344). We now also know that the rest of the mass of your unknown gas is that 'other' element because the initial question said that an excess of oxygen was used. So Hydrogen makes up 0.1344/1.31 = 10.26% of your unknown gas. You should be able to come up with the rest of the information you need.

thank you very much... this will be a probable test question.