How Accurate Is Gravimetric Analysis in Determining Mineral Composition?

[salman213]

The Title should actually be "Gravimetric Analysis"

srry about that but i can't change it now :(

1.
A 6.34-g geological sample, containing the mineral Boulangerite (Pb5Sb4S11) as the only lead-bearing compound, is subjected to chemical treatment in which the lead is quantitatively recovered as solid PbCl2. The mass of PbCl2 obtained is 3.80 g.
What was the percentage (by mass) of Pb5Sb4S11 in the sample?



2. n=m/MM
%yield= (actual yield over theoretical yield) * 100




3. Well I am guessing the equation is something like

Pb5Sb4S11 -------> PbCl2(s)

how to get cl in there or SbS out :S

mass is given as 3.8g


I have no clue
can someone can get me started atleast?

 

[chemisttree]

The Title should actually be "Gravimetric Analysis"

srry about that but i can't change it now :(

1.
A 6.34-g geological sample, containing the mineral Boulangerite (Pb5Sb4S11) as the only lead-bearing compound, is subjected to chemical treatment in which the lead is quantitatively recovered as solid PbCl2. The mass of PbCl2 obtained is 3.80 g.
What was the percentage (by mass) of Pb5Sb4S11 in the sample?



2. n=m/MM
%yield= (actual yield over theoretical yield) * 100

3. Well I am guessing the equation is something like

Pb5Sb4S11 -------> PbCl2(s)

how to get cl in there or SbS out :S

mass is given as 3.8g


I have no clue
can someone can get me started atleast?

If you knew the total number of moles of lead in the original sample could you use it to find out how many moles of $$Pb_5Sb_4S_{11}$$ were in the original sample?

 

[Blueshift5]

I would like to clarify that the topic being discussed here is gravimetric analysis, specifically the determination of percentage yield of a compound in a given sample. Stoichiometry is a fundamental concept in chemistry that relates the quantities of reactants and products in a chemical reaction, while yield refers to the actual amount of product obtained in a reaction compared to the theoretical amount.

To solve the given problem, we can use the equation %yield = (actual yield/theoretical yield) * 100. In this case, the actual yield is given as 3.80 g of PbCl2, while the theoretical yield can be calculated using the stoichiometry of the reaction. From the balanced equation, we can see that for every 1 mole of Pb5Sb4S11, 1 mole of PbCl2 is produced.

Using the molar mass of Pb5Sb4S11 (1063.5 g/mol) and PbCl2 (278.1 g/mol), we can calculate the theoretical yield of PbCl2 as 3.80 g * (278.1 g/mol / 1063.5 g/mol) = 0.99 g.

Plugging these values into the %yield equation, we get %yield = (3.80 g / 0.99 g) * 100 = 383.8%. This means that the actual yield of PbCl2 is 383.8% of the theoretical yield.

To find the percentage of Pb5Sb4S11 in the sample, we can use the same equation %yield = (actual yield/theoretical yield) * 100. In this case, the theoretical yield of Pb5Sb4S11 is equal to the actual yield of PbCl2, since they have a 1:1 ratio in the balanced equation.

Therefore, %yield = (6.34 g / 3.80 g) * 100 = 166.8%. This means that the percentage of Pb5Sb4S11 in the sample is 166.8%.

In conclusion, by using stoichiometry and the concept of yield, we can accurately determine the percentage of a compound in a given sample through gravimetric analysis.