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CHEMISTRY, M1 EQ-Bank 36

A mixture of sand and salt was provided to a group of students for them to determine its percentage composition by mass.

They added water to the sample before using filtration and evaporation to separate the components.

During the evaporation step, the students noticed white powder ‘spitting’ out of the basin onto the bench, so they turned off the Bunsen burner and allowed the remaining water to evaporate overnight.

After filtering, they allowed the filter paper to dry overnight before weighing. An electronic balance was used to measure the mass of each component to two decimal places.

The results were recorded as shown:

    • Mass of the original sand and salt mixture = 15.73 g
    • Mass of the filter paper = 0.80 g
    • Mass of the dried filter paper after filtering = 11.95 g
    • Mass of the empty evaporating basin = 33.50 g
    • Mass of the evaporating basin after evaporation = 36.60 g
  1. Calculate the percentage composition by mass of sand AND salt in the mixture.   (3 marks)

  2. Consider the following definition of validity
  3. Validity is the degree to which tests measure what was intended, or the accuracy of actions, data and inferences produced from tests and other processes.
  4. Use this definition of to assess the validity of the experiment.   (2 marks)

Show Answers Only

a.    % Sand = 70.88%

% Salt = 19.71%

b.   → The calculations show that the percentages do not add up to 100.

→ Some mass (salt) was observed to be lost during the experiment, and thus the mass of salt determined is lower than the true value and the results are not accurate.

→ Hence, experiment is not valid.

Show Worked Solution
a.     \(\text{Sand mass}\ \) \(\text{ = Mass of dried filter paper – Mass of filter paper}\)
    \(= 11.95-0.80 = 11.15\ \text{g}\)
\(\text{Salt mass}\) \(\text{ = Mass of dried filter paper – Mass of filter paper}\)  
  \(= 36.60-33.50 = 3.10\ \text{g}\)  

 

\(\text{% sand} = \left(\dfrac{\text{Mass of sand}}{\text{Mass of original mixture}}\right) \times 100= \left(\dfrac{11.15 \ \text{g}}{15.73 \ \text{g}}\right) \times 100 = 70.88\%\)

\(\text{% salt} = \left(\dfrac{\text{Mass of salt}}{\text{Mass of original mixture}}\right) \times 100 = \left(\dfrac{3.10 \ \text{g}}{15.73 \ \text{g}}\right) \times 100 = 19.71\%\)
 

b.   → The calculations show that the percentages do not add up to 100.

→ Some mass (salt) was observed to be lost during the experiment, and thus the mass of salt determined is lower than the true value and the results are not accurate.

→ Hence, experiment is not valid.

CHEMISTRY, M1 EQ-Bank 27

Silver oxide, when heated, decomposes according to the following chemical equation:

\(\ce{2Ag2O -> 4Ag + O2}\)

231.74 grams of silver oxide (\(\ce{Ag2O}\)) is heated, and the metallic silver (\(\ce{Ag}\)) residue is weighed and recorded in the table below.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{Ag2O} \rule[-1ex]{0pt}{0pt} & \text{231.74 grams} \\
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{Ag} \rule[-1ex]{0pt}{0pt} & \text{216.0 grams} \\
\hline
\end{array}

Calculate the percentage of oxygen (\(\ce{O2}\)), by weight, released from the silver oxide during this process.   (2 marks)

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\(\%\ce{O2} = 6.8\%\)

Show Worked Solution

\(\text{Initial mass of}\ \ce{Ag2O} = 231.74\ \text{grams}\)

\(\text{Mass of}\ \ce{Ag} = 216.0\ \text{grams}\)

\(\text{Mass of}\ \ce{O2} = 231.74-216.0 = 15.74\ \text{grams}\)

\(\%\ce{O2}\ \text{released}\ = \left(\dfrac{15.74}{231.74} \times 100\%\right) = 6.8\%\)

CHEMISTRY, EQ-Bank 26

Calcium carbonate, when heated, decomposes according to the following chemical equation:

\(\ce{CaCO3 -> CaO(s) + CO2(g) }\)

60.0 grams of calcium carbonate is heated and the calcium oxide residue produced is weighed and recorded in the table below.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{CaCO3} \rule[-1ex]{0pt}{0pt} & \text{60.0 grams} \\
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{CaO} \rule[-1ex]{0pt}{0pt} & \text{33.6 grams} \\
\hline
\end{array}

Determine the percentage of carbon dioxide, by weight, released from the calcium carbonate during this process.   (2 marks)

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\(\%\ce{CO2} = 44.0\%\)

Show Worked Solution

\(\text{Mass of}\ \ce{CaCO3} = 60.0\ \text{grams}\)

\(\text{Mass of}\ \ce{CO2} = 60.0-33.6 = 26.4\ \text{grams}\)

\(\%\ce{CO2}\ \text{released}\ = \dfrac{26.4}{60.0} \times 100\% = 44.0\%\)

CHEMISTRY, M1 EQ-Bank 25

15.00 grams of Silver(\(\text{I}\)) oxide is heated until it decomposes and the silver produced is weighed and recorded in the table below.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{Ag2O} \rule[-1ex]{0pt}{0pt} & \text{15.00 grams} \\
\hline
\rule{0pt}{2.5ex} \text{Mass of}\ \ce{Ag} \rule[-1ex]{0pt}{0pt} & \text{13.96 grams} \\
\hline
\end{array}

Determine the percentage of oxygen, by weight, in the sample of Silver(\(\text{I}\)) oxide.   (2 marks)

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\(\%\ce{O} = 6.93\%\)

Show Worked Solution

\(\text{Mass of}\ \ce{Ag} = 13.96\ \text{grams}\)

\(\text{Mass of}\ \ce{O} = 15.00-13.96 = 1.04\ \text{grams}\)

\(\%\ce{O} = \dfrac{1.04}{15.00} \times 100\% = 6.93\%\)

CHEMISTRY, M1 EQ-Bank 5

Calculate the mass of \(\ce{H}\) in 15.2 grams of \(\ce{H2O}\). Give your answer in grams correct to two decimal places.   (2 marks)

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\(1.70\ \text{grams}\)

Show Worked Solution

\(\ce{MM(H2O) = 2 \times 1.008 + 16.00 = 18.016\ \text{g mol}^{-1}}\)

\(\ce{\% of H = \dfrac{2.016}{18.016} \times 100 = 11.19\%}\)

\(\ce{m(H)\ \text{within 15.2 grams}\ H2O = 11.19\% \times 15.2 = 1.70\ \text{grams (to 2 d.p.)}} \)

CHEMISTRY, M1 EQ-Bank 4

Calculate the mass of \(\ce{H}\) within 100 grams of the compound \(\ce{H2CO3}\). Give your answer in grams correct to two decimal places.   (2 marks)

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\(3.25\ \text{grams}\)

Show Worked Solution

\(\ce{MM(H2CO3) = 2 \times 1.008 + 12.01 + 3 \times 16.00 = 62.026\ \text{g mol}^{-1}}\)

\(\ce{\% of H = \dfrac{2.016}{62.026} \times 100 = 3.25\%}\)

\(\ce{m(H)\ \text{within 100g}\ H2CO3 = 3.25\% \times 100 = 3.25\ \text{grams}} \)

CHEMISTRY, M1 EQ-Bank 3

Calculate the mass of \(\ce{O}\) in 8.4 grams of \(\ce{CO2}\). Give your answer in grams correct to two decimal places.   (2 marks)

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\(6.11\ \text{g}\)

Show Worked Solution

\(\ce{MM(CO2) = 12.01 + 2 \times 16.00 = 44.01\ \text{g mol}^{-1}}\)

\(\ce{\% of O = \dfrac{32.00}{44.01} \times 100 = 72.71\%}\)

\(\ce{m(O\ \text{in 8.4 g})}\) \(= 8.4 \times 72.71\% \)  
  \(=6.11\ \text{g (2 d.p.)}\)