Unknown number Mass of crucible and cover (g) Mass of crucible, cover, and mixtu

ID: 1033280 • Letter: U

Question

Unknown number Mass of crucible and cover (g) Mass of crucible, cover, and mixture (g) Mass of crucible, cover, and residue after reaction with HCI (g) Run 1 2? Sp. 8333 e with HCI Run 2 9226 & 3230 8.312 1st weighing 2nd weighing 3rd weighing, if necessary 4th weighing, if necessary Instructor's initias CALCULATIONS Table 5.2. Calculation of percent NaHCO, and Na,CO, in an unknown mixture Unknown number Run 1 Run 2 Mass of unknown mixture used (g) o. 499 911 O.S?ro Mass of NaCI formed (g) Mass of NaHCO, in unknown mixture (g) Mass of Na, CO, in unknown mixture (g) Percent of NaHCO, by mass in unknown mixture (%) Percent of Na,CO, by mass in unknown mixture (%) Average percent by mass of other elements in unknown sample (%) EXPERIMENT 5

Explanation / Answer

Run 1

Run2

Mass of unknown mixture used (g)

0.5000

0.4999

Mass of NaCl formed (g)

0.1020

0.1026

Mass of NaHCO3 in the unknown mixture (g)

0.0489076

0.0491953

Mass of Na2CO3 in the unknown mixture (g)

0.1234106

0.1241366

Percent of NaHCO3 by mass in unknown mixture (%)

9.78

9.84

Percent of Na2CO3 by mass in unknown mixture (%)

24.68

24.83

Average percent by mass of other elements in unknown sample (%)

65.54

65.33

Example calculations for Run 1

M.W. of NaCl = 58.40 g/mol

M.W. of NaHCO3 = 84.006 g/mol

M.W. of Na2CO3 = 105.988 g/mol

On adding excess HCl following reaction happens

Mass of NaCl formed = 28.3330 – 28.2310 = 0.1020 g

i.e. 1 mole of NaCl formed from 1 mole of NaHCO3 and 2 mole of NaCl formed from 1 mole of Na2CO3

Moles of NaCl formed = 0.1020/58.40 = 0.001746575

So the moles of NaHCO3 = 1/3 (0.001746575) = .00058219178

Mass of NaHCO3 = .00058219178 x 84.006 = 0.0489076 g

So the moles of Na2CO3 = 2/3 (0.001746575) = 0.0011643833

Mass of Na2CO3 = 0.0011643833 x 105.988 = 0.1234106 g

Percent of NaHCO3 by mass in unknown mixture (%) = (0.0489076 / 0.5000) x 100 = 9.78 %

Percent of NaHCO3 by mass in unknown mixture (%) = (0.1234106 / 0.5000) x 100 = 24.68 %

Average percent by mass of other elements in unknown sample (%) = 100 -24.68 -9.78 = 65.54 %