A compound that is composed of carbon hydrogen and oxygen contains 70.6 C 5.9 H and 23.5 O by mass. The molecular weight of the compound is 136 amu. What is the molecular formula?

Answer 1

C8H8O2 = 136amu

Answer 2

The gram molecular mass of carbon dioxide is 44.01, and the gram molecular mass of water is 18.015. Therefore, 3.84g of carbon dioxide constitutes 3.84/44.01 or about 0.08725 mole of carbon dioxide, and 1.05g of water constitutes 1.05/18.015 or about 0.05828 mole of water or 0.1157 gram atoms of hydrogen, since each molecule of water contains two hydrogen atoms. Therefore, the ratio of the numbers of hydrogen atoms to the numbers of carbon atoms is about 0.1157/0.08725 or about 1.34. The smallest whole numbers with this approximate ratio are 4 and 3.

The only products of complete combustion of a hydrocarbon or oxyhydrocarbon are carbon dioxide and water. If the compound contained no oxygen, the number of oxygen atoms needed from the atmosphere would be two for each atom of carbon plus one half for each atom of hydrogen, for a total of 8 (for 3 carbon atoms and 4 hydrogen atoms) for each "empirical molecule", and the ratio of the mass of the initial compound and the masses of the products of complete combustion would be the same as the ratio of the mass of 8 oxygen atoms plus 3 carbon atoms plus 4 hydrogen atoms to the mass of 3 carbon atoms plus 4 hydrogen atoms alone, or {[8(15.9994)] + [3(12.011)] + 4(1.008)]}/{[3(12.011)] + 4(1.008)]} or about 4.196. The actual mass gain, however, from the starting compound to the combustion products is only (3.84 + 1.05) - 2.56 or about 2.33. Therefore, the mass in grams of oxygen already present in the compound, designated Oc, must satisfy the relationship (Oc + 2.33)/2.56 = 4.196, or Oc = about 1.87. 1.87/15.9994 or about 0.1169. The ratio of the number of atoms of internal oxygen to the number of atoms of internal carbon is therefore 0.187/0.08725, or about 2. The most probable empirical formula is C3H4O6.

Answer 3

27.8 g CO2 x 1 mole CO2/44 g x 1 mole C/mole CO2 = 0.632 moles C = 7.58 g C19.9 g H2O x 1 mole H2O/18 g x 2 mole H/mole H2O = 2.21 moles H = 2.21 g H

grams O = 25.0 g - 7.58 g - 2.21 g = 15.21 g O = 0.951 moles O

Mole ratio C:H:O = 0.632 : 2.21 : 0.951 or dividing all by 0.632 to get 1 : 3.5 : 1.5

To get all whole numbers, multiply all by 2 to get C2H7O3 as the empirical formula

Answer 4

i dont k

now

Answer 5

C2h5

Answer 6

0.25mol

Answer 7

C2H5