Question

In: Chemistry

A 9.59 g sample of calcium sulfide was decomposed into its constituent elements, producing 5.33 g...

A 9.59 g sample of calcium sulfide was decomposed into its constituent elements, producing 5.33 g of calcium and 4.26 g of sulfur. Which of the statements are consistent with the law of constant composition (definite proportions)? The mass percentage of calcium plus the mass percentage of sulfur in every sample of calcium sulfide equals 100%. The ratio of calcium to sulfur will vary based on how the sample was prepared. The mass ratio of Ca to S in every sample of calcium sulfide is 1.25. Every sample of calcium sulfide will have 5.33 g of calcium. Every sample of calcium sulfide will have 55.6% mass of calcium

Solutions

Expert Solution

Consistent statements are:

The mass percentage of calcium plus the mass percentage of sulfur in every sample of calcium sulfide equals 100%.

The mass ratio of Ca to S in every sample of calcium sulfide is 1.25.

Every sample of calcium sulfide will have 55.6% mass of calcium.

---

The law of constant composition states that the proportion of masses of elements present in a compound is fixed no matter what its source is or whichever way the compound is prepared.

Consistent statement 1 explanation:

In calcium sulfide or CaS, the mass percentage of calcium plus the mass percentage of sulfur in every sample of calcium sulfide equals 100%.

Mass of calcium = 5.33 g

Mass of the sample = 9.59 g

Hence, % Ca = (5.33 g x 100)/9.59 g = 55.6 %

Mass of sulfur = 4.26 g

Mass of the sample = 9.59 g

Hence, % S = (4.26 g x 100)/9.59 g = 44.4 %

Thus, % Ca + % S = (55.6 + 44.4) % = 100 %

Consistent statement 2 explanation:

In calcium sulfide or CaS, the mass ratio of Ca to S in every sample of calcium sulfide is 1.25.

ratio of %Ca : % S = 55.6 : 44.4 = 1.25 : 1

Consistent statement 3 explanation:

In calcium sulfide or CaS, every sample of calcium sulfide will have 55.6% mass of calcium.

Mass of calcium = 5.33 g

Mass of the sample = 9.59 g

Hence, % Ca = (5.33 g x 100)/9.59 g = 55.6 %

Incorrect statement:

The ratio of calcium to sulfur will vary based on how the sample was prepared.

It is incorrect because, in 1 one molecule of CaS, there will always be 1 Ca and 1 S, no matter how the CaS is prepared.

Incorrect statement:

Every sample of calcium sulfide will have 5.33 g of calcium.

This is incorrect because not every sample will have 5.33 g of Ca as long as the ratio of masses of Ca and S is 1.25. Thus a sample of CaS can contain any arbitrary mass of Ca.

For example, a sample of CaS can contain 10 g of Ca. Under this condition, the CaS must contain 10/1.25 = 8 g of S.


Related Solutions

A 1.912-g sample of calcium chloride is decomposed into its constituent elements and found to contain...
A 1.912-g sample of calcium chloride is decomposed into its constituent elements and found to contain 0.690 g Ca and 1.222 g Cl. Calculate the mass percent composition of Ca and Cl in calcium chloride.
A sample of XeO4 was fully decomposed into its elements, and the resulting gas was collected...
A sample of XeO4 was fully decomposed into its elements, and the resulting gas was collected over water at 29°C. If the total pressure of the collected gas was 750 torr, and the vapor pressure of water at 29°C is 30.0 torr, what was the partial pressure of O2 gas in the flask?
Calcium reacts with sulfur (S8) forming calcium sulfide. What is the theoretical yield (g) of CaS(s)...
Calcium reacts with sulfur (S8) forming calcium sulfide. What is the theoretical yield (g) of CaS(s) that could be prepared from 1.83 g of Ca(s) and 3.35 g of sulfur(s)? Enter your answer in decimal format with two decimal places and no units.
A 2.95 g sample of an unknown chlorofluorocarbon is decomposed and produces 579 mL of chlorine...
A 2.95 g sample of an unknown chlorofluorocarbon is decomposed and produces 579 mL of chlorine gas at a pressure of 756 Torr and a temperature of 298 K. Part A What is the percent chlorine (by mass) in the unknown chlorofluorocarbon?
4. A 0.3634 g sample of ore containing calcium is dissolved and the calcium ions are...
4. A 0.3634 g sample of ore containing calcium is dissolved and the calcium ions are precipitated as CaC2O4. The CaC2O4 precipitate is filtered, washed and redissolved in acid. The pH of the resulting Ca2+ solution is adjusted. A 25.00 mL volume of 0.0376 M EDTA is added. The excess EDTA is back-titrated to the end point, requiring 23.08 mL of 0.0188 M Mg2+ solution. a) Write a balanced chemical equation for the EDTA reactions that occur as part of...
The hydrogen sulfide in a 82.868 g sample of crude petroleum was removed by distillation and...
The hydrogen sulfide in a 82.868 g sample of crude petroleum was removed by distillation and collected in a solution of CdCl2. The precipitated CdS was then filtered, washed and ignited to create CdSO4. Calculate the percentage of H2S in the sample if 0.192 grams of CdSO4 was formed.
Calculate the enthalpy change for producing 215 g AlCl3(s) from its elements in their natural state...
Calculate the enthalpy change for producing 215 g AlCl3(s) from its elements in their natural state using the following thermochemical equations.    Given: ∆Hºrxn 2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2 -1049 HCl(g) → HCl(aq)    -73.5 ½H2(g) + ½Cl2 → HCl(g) -92.5 (∆Hºf) AlCl3(s) → AlCl3(aq) -323
Calculate the enthalpy change for producing 215 g AlCl3(s) from its elements in their natural state...
Calculate the enthalpy change for producing 215 g AlCl3(s) from its elements in their natural state using the following thermochemical equations.    Given: ∆Hºrxn 2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2 -1049 HCl(g) → HCl(aq)    -73.5 ½H2(g) + ½Cl2 → HCl(g) -92.5 (∆Hºf) AlCl3(s) → AlCl3(aq) -323
A sample of NaN3(s) fully decomposed into Na(s) and N2(g), and the resulting gas was collected...
A sample of NaN3(s) fully decomposed into Na(s) and N2(g), and the resulting gas was collected over water at 52°C. The collected gas occupied 2.0 L at a pressure of 690 torr. If the vapor pressure of water is 100 torr at 52°C, what mass of NaN3(65.02 g/mol) was reacted? (1) 2.5 g (2) 2.9 g (3) 3.4 g (4) 3.8 g (5) 4.4 g
the density of the sample is 0.9977 g/mL and the molar mass of calcium carbonate is...
the density of the sample is 0.9977 g/mL and the molar mass of calcium carbonate is 100.0 g/mol. 2. Calculate the number of moles of calcium ions present in a 50.00 mL water sample that has a hardness of 75.0 ppm (hardness due to CaCO3). 3. If the 50.00 mL sample from problem 2 above was titrated with a 0.00500 M EDTA, what volume (in milliliters) of EDTA solution would be needed to reach the endpoint? If someone could show...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT