Upon decomposition, one sample of magnesium fluoride produced 1.65 kg of magnesium and 2.56 kg of fluorine. a second sample produced 1.32 kg of magnesium. part a how much fluorine (in grams) did the second sample produce?

Answer : The mass of fluorine in second sample produces, 2.048 grams

Law of definite proportion states that in a chemical compound, the components of the element are in fixed ratio.

The given chemical compound is magnesium fluoride. In the magnesium fluoride the components of the element magnesium and fluorine are in fixed ratio that is, 1 : 1

= mass of magnesium in sample 1 = 1.65 g

= mass of fluorine in sample 1 = 2.56 g

= mass of magnesium in sample 2 = 1.32 g

= mass of fluorine in sample 2 = x

Now put all the give values in the above relation, we get

Therefore, the mass of fluorine in second sample produces, 2.048 grams

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o

the answers are

c and c