Answer to Question #185941 in General Chemistry for ary

Question

Magnesium can be determined gravimetrically by precipitating as NH4MgPO4. A 1.291 g sample was dissolved and the magnesium precipitated as NH4MgPO4. The precipitated was collected, rinsed and ignated to constant weight as Mg2P2O7, yielding 0.1258 g. Calculate the %w/w Mg in the sample.

Answer 1

When NH4MgPO4 is ignited , the equation for the igniiton is provided below :

Heat

2 NH4MgPO4 1 2 NH3 (g) + H2O (g) + Mg2P2O7 (s)

The mass of Mg2P2O7 obtained is 0.1258 grams.

Molar mass of Mg2P2O7 = 2 × 24.3 + 31 × 2 + 7 × 16 = 48.6 + 62 + 112 = 222.6 grams/mol

.

Moles of Mg2P2O7 produced = 0.1258 / 222.6 g/mol = 5.65 × 10-4 moles

2 moles of NH4MgPO4 produce 1 mole of Mg2P2O7 as per balanced equation .

11.3 × 10-4 moles of NH4MgPO4 will produce 5.65 × 10-4 moles of Mg2P2O7 as per balanced equation .

Same number of moles of Mg will be present in NH4MgPO4 as the total number of moles of NH4MgPO4.

Moles of Mg in NH4MgPO4 = 11.3 × 10-4 moles

Molar mass of Magnesium = 24.3 g/mol

Total Magnesium present in the sample is in the precipitate (NH4MgPO4 ) :

Mass of Magnesium in NH4MgPO4 = 11.3 × 10-4 × 24.3 = 0.0274 grams

Total mass of sample containing Magnesium = 1.291 grams

%w/w of Magnesium in the sample = Mass of Magnesium / Total mass of the sample = (0.0274 / 1.291) × 100 = 2.12 %

2.12 % w/w of Magnesium is present in 1.291 grams of sample.