Question

Water is added to 4.267 g UF6. The only products are 3.730 g of a solid containing only uranium, oxygen and fluorine and 0.970 grams of gas. The gas is 95.0% by mass fluorine and the remainder is hydrogen.
(a) From the data, determine the molecular formula of the gas (same as the empirical formula)
(b) What is the mass of fluorine in UF6?
(c) What is the mass of uranium in UF6?
(d) What is the mass of fluorine in the solid product?
(e) Determine the molecular formula (same as empirical) for the solid product the write a balanced chemical equation for the reaction.

Answer 1

Answer:

a) HF

b) $m_F=1.382g F$

c) $m_U=2.885g U$

d) $m_{F-solid}=0.4605g$

e) $UF_2O_2$

Explanation:

a) Mass of the gas: 0.970 g

Knowing that 95% is fluorine:

$m_F= 0.95*0.97g= 0.9215g$

and the rest is H:

$m_H=0.97g-0.9215g=0.0485g$

Expressing in mol:

$n_F=\frac{0.9215g}{19g/mol}=0.0485mol$

$n_H=\frac{0.0485g}{1g/mol}=0.0485mol$

As can be seen, the gas has 1 mol of F per mol of H, so the molecular formula is: HF

b) First we need to calculate the mass ratio between F and U. For 1 mol of UF6:

$m_U=1mol*238 g/mol=238g$

$m_F=6mol*19 g/mol=114g$

This means that for each 352 g of UF6 (114g +238g) there are 238 g of U and 114  g of F.

In the sample:

$m_F=4.267 g * \frac{114g F}{352 g}$

$m_F=1.382g F$

c) Now for the uranium:

$m_U=4.267 g * \frac{238g U}{352 g}$

$m_U=2.885g U$

d) We know that:

$m_{F-solid}=m_{F-total} - m_{F-gas}$

$m_{F-solid}=1.382g - 0.9215g=0.4605g$

e) For the molecular formula:

Fluorine: $n_{F-solid}=\frac{0.4605gF}{19gF/mol}=0.0242mol$

Uranium: $n_{U}=\frac{2.885gU}{238gU/mol}=0.0121mol$

Oxygen came from the water so the moles are half of the H moles:

$n_{O}=\frac{0.0485mol}{2}=0.0242mol$

Dividing by 0.0121 mol:

$n_{F-solid}=2$

$n_{U}=1$

$n_{O}=2$

The molecular formula: $UF_2O_2$