Answer to Question #182857 in Chemistry for Kristagay Tulloch

Question

7.A mineral found in a mine site near Bancroft, ON is identified as “Barite”. A geochemist’s analysis finds that it consists of 21.93% barium, 5.12% sulfur and 10.24% oxygen. (7 marks)

a.Determine the empirical formula

b.Further analysis of the mineral determined the molecular mass of the compound to actually be, M°MF = 233.40 g/mol. Find the molecular formula for this compound.

8.What is the mole ratio for lithium hydroxide to hydrogen gas in the following reaction?

2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g)   (1mark)

9.Aluminum carbide, Al4C3(s), is a yellow powder that reacts with water, H2O(ℓ), to produce aluminum hydroxide, Al(OH)3(s), and methane, CH4(g).

1. Write a balanced chemical equation for the reaction and (2)
2. Determine the mass of water required to react with 14.0 g of aluminum carbide. (5)

Answer 1

Solution:

(7):

Barium = Ba; sulfur = S; oxygen = O.

Since the scale for percentages is 100, it is most convenient to calculate the mass of elements present in a sample weighing 100 g. Taking this into account, the mass percentages provided may be more conveniently expressed as fractions:

21.93% Ba = 21.93 g Ba / 100 g sample

5.12% S = 5.12 g S / 100 g sample

10.24% O = 10.24 g O / 100 g sample

Using the molar masses and masses of each elements to convert their to moles:

For Ba: (21.93 g Ba) × (1 mol Ba / 137.327 g Ba) = 0.1597 mol Ba

For S: (5.12 g S) × (1 mol S / 32.065 g S) = 0.1597 mol S

For O: (10.24 g O) × (1 mol O / 15.999 g O) = 0.6400 mol O

Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the three:

0.1597 mol Ba / 0.1597 = 1

0.1597 mol S / 0.1597 = 1

0.6400 mol O / 0.1597 = 4

This means that: BaSO4 - empirical formula for this compound

Empirical formula mass = Ar(Ba) + Ar(S) + 4×Ar(O) = 137.327 + 32.065 + 4×15.999 = 233.388 (g/mol)

molar mass / empirical formula mass = n formula units/molecule

(233.40 g/mol) / (233.388 g/mol) = 1.00

This means that: BaSO4 - molecular formula for this compound

(8):

Balanced chemical equation:

2Li(s) + 2H₂O(l) → 2LiOH(aq) + H₂(g)

According to the equation above: n(Li) = n(H₂O) = n(LiOH) = 2×n(H₂)

Hence,

n(LiOH) / n(H₂) = 2

Moles of LiOH : Moles of H₂ = 2 : 1

(9):

Balanced chemical equation:

Al₄C₃(s) + 12H₂O(l) → 4Al(OH)₃(s) + 3CH₄(g)

According to the equation above: n(Al₄C₃) = n(H₂O)/12

The molar mass of Al₄C₃ is 143.9585 g/mol.

The molar mass of H₂O is 18.0153 g/mol.

Hence,

(14.0 g Al₄C₃)×(1 mol Al₄C₃/143.9585 g Al₄C₃)×(12 mol H₂O/1 mol Al₄C₃)×(18.0153 g H₂O/1 mol H₂O) = 21.024 g H₂O = 21.0 g H₂O

Answers:

7a) The empirical formula is BaSO₄

7b) The molecular formula is BaSO₄

8) Moles of LiOH : Moles of H₂ = 2:1

9.1) Balanced chemical equation: Al₄C₃(s) + 12H₂O(l) → 4Al(OH)₃(s) + 3CH₄(g)

9.2) 21.0 g H₂O