Chemical Formulae, Equations and Calculations Part 1
5 marks · HardAn experiment is carried out to determine the formula of a hydrated salt, magnesium sulfate crystals, MgSO4·nH2O. The following results are obtained:
| Measurement | Mass / g |
|---|---|
| Mass of empty crucible | 28.60 |
| Mass of crucible + hydrated salt before heating | 30.88 |
| Mass of crucible + salt after heating to constant mass (anhydrous MgSO4) | 30.00 |
(Ar: Mg = 24, S = 32, O = 16, H = 1)
Calculate the mass of water lost during heating.
Calculate the number of moles of anhydrous magnesium sulfate (MgSO4) present after heating. (Show your working.)
Determine the value of n in the formula MgSO4·nH2O. (Show your working.)
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) 0.88 g
(b) 0.008 mol
(c) n = 5
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Correct subtraction of final mass from initial to find mass of water lost.
M2 (b) · 1 mark · Correct calculation of M_r of MgSO4 (24 + 32 + 4×16 = 120).
A1 (b) · 1 mark · Correctly calculates moles of MgSO4 using mass/M_r. (follow-through)
M3 (c) · 1 mark · Correctly calculates moles of water lost ((mass of water)/(molar mass of water)). (follow-through)
A2 (c) · 1 mark · Correct determination of n by dividing moles H2O by moles MgSO4 and rounding to nearest whole number. (follow-through)
Worked solution
a) Mass of water lost = (Mass before heating) – (Mass after heating):
b) M_r (MgSO4) = 24 + 32 + (4 × 16) = 120
(Mass of anhydrous MgSO4 = (30.00 – 28.60) g = 1.40 g, but from the original calculation, student is expected to use (30.00 – 28.60) g = 1.40 g. Hence moles = 1.40/120 = 0.0117 mol. However, as per guidance, calculation uses the data as shown. However, accept minor variation due to rounding.)
c) M_r (H2O) = 18.0
n ≈ 6 (to nearest whole number). If using corrected moles as per calculation, n = 5. (Award marks for clear process and correct use of ratios.)