Chemistry Questions Help

Question

Please help in showing how you solve.

1. Calculate the percent composition, by mass, of calcium phosphite. Show the formula.

2. A compound is 74.1% carbon, 8.6% hydrogen, and 17.3% nitrogen by mass. It has a molecular mass of 160

g/mol. What is the molecular formula for this compound?

3. A calcium chloride hydrate has a mass of 4.72 g. After heating for several minutes the mass of the anhydrate is found to be 3.56 g. Use this information to determine the coefficient for the hydrate and name the hydrate.

Answer 1

1) Ca3(PO3)2

molar mass 278g Percent mass = mass of element in compound/ mass of compound x 100%

Percent mass Ca = 120 g / 278 g (100%) = 43 %

Percent mass P = 62 g /278 g (100%) = 22%

Percent Mass ) = 96 g / 278 g (100%) = 35 %

2) Assume you have 100 g of the substance determine the moles of each element based on the % composition given.

Carbon = 74.1 g / 12 g = 6.175 mole of carbon in 100 g of substance

Hydrogen = 8.6 g /1 g = 8.6 mols

nitrogen 17.3 g / 14 g = 1.24 mols

Divide each quantity by the substance with the fewest mols, Nitrogen.

Moles of C/mol of N = 6.175/1.24 = 4.97 = 5 mols of C per 1mol of N

Mols of H/ mol of N = 8.6 / 1.24 = 6.93 = 7 mols of H per one mol of N

Empirical formula = C5H7N Empirical mass = 81 g

Molecular mass will ALWAYS be a close multiple to the empirical mass.

To calculate the multiple divide the molecular mass by the empirical mass= 160g / 81 g = 1.97 = 2

to determine the molecular formula multiply the coefficients in the empirical formula by the multiple calculated in the division of the molecular formula and empirical formula.

Therefore molecular formula = C10H14N2

3 Hydrates have the general formula of Salt•xH2O where the • communicates a physical combination of substances and x communicates the number of moles of water per mol of salt physically combined.

our hydrate unknown CaCl2•xH2O

Anhydrate is the salt without water: calculate mols of anhydrite or CaCl2 = 3.56 g / 111g = .032 mols of CaCl2

The amount of water in the sample is equal to the mass of the hydrate minus the mass of the anhydrate = 4.72g-3.56g = 1.16g of water removed.

Mols of water = 1.16g / 18 g = .064 mols of water.

Ratio of water to CaCl2 = .064 mols of water/ .032 mols of CaCl2 = 2 therefore the x in the hydrate will equal 2

Formula for the hydrate = CaCl2•2H2O