1. A sample of a compound of Cl and O reacts with excess H 2 to give 0.233 g of
ID: 840900 • Letter: 1
Question
1. A sample of a compound of Cl and O reacts with excess H2 to give 0.233 g of HCl and 0.403 g of H2O. Determine the empirical formula of the compound.
2. The atomic mass of element X is 33.42 amu. A 27.22 g sample of X combines with 84.10 g of another element Y to form XY. Calculate the atomic mass of Y.
3. Carbohydrates are compounds containing carbon, hydrogen and oxygen in which hydrogen to oxygen ratio if 2:1. A certain carbohydrate contains 40.0 percent carbon by mass. Calculate the empirical and molecular formulas of the compound if the approximate molar mass if 178 g.
4. Analysis of a metal chloride XCl3 shows that it contains 67.2 percent Cl by mass. Calculate the molar mass of X and identify the element.
5. Hydrazine (N2H4) and hydrogen peroxide (H2O2) have been used as rocket propellants. They react to the following equation:
7 H2O2 + N2H4
Explanation / Answer
1.moles HCl = 0.233/36.461 g/mol=0.00639
moles H2O = 0.403 / 18.02 g/mol=0.0224
0.0224 / 0.00639 = 3.5 => O
Cl O 3.5
to get whole numbers multiply by 2
Cl2O7
2.)Moles X = 27.22 g / 33.42 = 0.8145
= moles Y
Molar mass Y = 84.10 g / 0.8145 = 103.3 amu
3.If the molar mass is 178 g and the substance is 40% carbon by mass, the molar mass of the carbon in the substance is 178*0.4 = 71.2. The molar mass of a carbon atom is 12, and 71.2 / 12 = 5.93, so we can safely assume there are 6 carbon atoms in the compound. The remaining molar mass is 178 - 71.2 = 106.8 g. A single unit of H2O has a molar mass of 1*2 + 16 = 18, and 106.8 / 18 = 5.9333333333, so we can again say that there are 6 such units. That means we have C6H12O6, the molecular formula. The empirical formula requires division by the greatest common factor, in this case 6, leaving us with CH2O.
4) We consider 100 g of this compound
Mass Cl = 67.2 g
Moles Cl = 67.2 / 35.453 = 1.89
The formula is XCl3
The ratio between X and Cl is 1 :3
1 : 3 = x : 1.89
moles X = 0.63
Mass X = 100 - 67.2 = 32.8 g
amu of X = 32.8 g / 0.63 = 52.0
X is Cr
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