Question

1. Calcium oxalate monohydrate, CaC2O4•H2O, is stable up to ~100˚C, but will release the incorporated water molecule when heated to 300˚C, according to the following reaction:

CaC2O4•H2O(s) → CaC2O4(s) + H2O(g)

If a 4.1754 g sample of calcium oxalate monohydrate is heated to 300˚C, what should be the mass (g) of the resulting calcium oxalate?

2. An unknown sample of ore with a mass of 506.1 g is allowed to react with excess magnesium and ammonia, NH3, in order to analyze for phosphate, PO43-, according to the following precipitation reaction:

2PO43-(aq) + 2Mg2+(aq) + 2NH3(aq) + 2H2O(l) → Mg2(P2O7)(s) + 2NH3(aq) + 2OH-(aq) + H2O(l)

After the precipitated Mg2(P2O7) is carefully rinsed and dried, the mass is determined to be 0.09870 mg.
Assuming all phosphorous originally present in the ore was in the form of phosphate, what was the original mass (μg) of phosphorous in the ore sample?

3. An unknown organic compound was produced by students in the O-chem lab. According to their records, this compound should only contain carbon and hydrogen atoms. A combustion analysis is performed on the sample.

When 0.1727 g of the sample is burned in the presence of excess oxygen, the magnesium perchlorate cartridge increases 0.24889 g and the ascarite cartridge increases 0.53094 g.

What is the percent C (w/w) in the sample?

4. What is the percent H (w/w) in the sample?

5. What is the empirical formula for the unknown compound?

Answer 1

CaC2O4•H2O(s) → CaC2O4(s) + H2O(g)

As per the reaction, one mole of Calcium monohydrate crystal gives rise to one mole of calcium oxalate

Mole =mass/molecular weight for one mole, mass= molecular weight

molecular weigt og Calcium oxalate monohydrate =40+2*12+4*16+18=146 gms

molecular weight of Calcium oxalate ( excluding water ) =128 gms

146 gms givve rise to 128gms

4.1754 gms gives 4.1754*128/146= 3.06gms of Calcium oxalate

II

2PO43-(aq) + 2Mg2+(aq) + 2NH3(aq) + 2H2O(l) → Mg2(P2O7)(s) + 2NH3(aq) + 2OH-(aq) + H2O(l)

As per the reaction , 2moles of PO43- gives one mole of Mg2P2O7.

molecular weight og PO43- =31+4*16= 95 ratio of PO4-3 to P =95/31 =3.064

Molecular weight of Mg2P2O7= 2*12+31*2+7*16=222

2*95 gms (as per stoichiometry) of PO43- gives 222 gms of Mg2P2O7.

222 gms are obtained from 190 gms of PO43-

0.09870 mg. requires 0.09870*190/222 mg = 0.084473mg =84.473ug of PO43-

Phosphorous in the original minearl = 84.473/3 = 28.15766 ug

III Ascarite absorbs CO2 and perchlorate absorbs water. The rise in mass of Ascarite is due to CO2 which is 0.53094gms

Amount of CO2 generated due to combustion 0.53094 gms

The reaction is C+O2---> CO2

44gms of CO2 is obtained from 12gms of C

0.53094 is obtained from 0.53094*12/44= 0.144802gms of C

the increase in mass of perchlorate is 0.24889, mass of water =0.24889gm

The reaction is H2+1/2O2---> H2O

18 gms water requires 2 gms of H2

0.24889 gms of water requires 0.24889*2/18= 0.027654 gms of Hydrogen

Total mass of the hydrocaron containing C and H =0.144802+0.027654= 0.172456 gms

Pecent C :( 0.144802/0.172456)*100= 83.9%

Percent H =100-83.9 =15.1%

Mass ratio of C : H = 0.144802 : 0.027654

molar ratio = 0.144802/12 : 0.027654/1 = 0.012067 : 0.027654 = 1: 0.027654/0.012067= 1:2

the empirical formula is CH2