501. A hydrocarbon has empirical formula \( \mathrm{CH_2} \) and molar mass \(84\,\text{g mol}^{-1}\). It decolourises bromine water. What molecular formula is most likely?
ⓐ. \( \mathrm{C_6H_{12}} \)
ⓑ. \( \mathrm{C_3H_6} \)
ⓒ. \( \mathrm{C_7H_{14}} \)
ⓓ. \( \mathrm{C_6H_6} \)
Correct Answer: \( \mathrm{C_6H_{12}} \)
Explanation: \( \textbf{Empirical formula:} \)
\[
\mathrm{CH_2}
\]
\( \textbf{Empirical formula mass:} \)
\[
12+2(1)=14
\]
\( \textbf{Given molar mass:} \)
\[
84\,\text{g mol}^{-1}
\]
\( \textbf{Multiplier:} \)
\[
\frac{84}{14}=6
\]
\( \textbf{Molecular formula from empirical formula:} \)
\[
\mathrm{C_6H_{12}}
\]
\( \textbf{Bromine-water clue:} \) Decolourisation of bromine water is consistent with unsaturation such as a \( \mathrm{C=C} \) bond.
\( \textbf{Formula consistency:} \) \( \mathrm{C_6H_{12}} \) can represent an alkene or a cycloalkane, but the bromine-water clue supports an alkene-type unsaturation in this context.
\( \textbf{Final answer:} \) The molecular formula is \( \mathrm{C_6H_{12}} \).
The empirical formula gives the atom ratio, while the molar mass fixes the actual multiple of that ratio.
502. A sample gives the following data:
| Measurement | Value |
|---|
| Mass of organic compound | \(0.900\,\text{g}\) |
| Mass of \( \mathrm{CO_2} \) | \(1.320\,\text{g}\) |
| Mass of \( \mathrm{H_2O} \) | \(0.540\,\text{g}\) |
| Nitrogen percentage | \(15.56\%\) |
If the compound contains only \( \mathrm{C} \), \( \mathrm{H} \), \( \mathrm{N} \), and \( \mathrm{O} \), what mass of oxygen is present in the sample?
ⓐ. \(0.140\,\text{g}\)
ⓑ. \(0.360\,\text{g}\)
ⓒ. \(0.500\,\text{g}\)
ⓓ. \(0.340\,\text{g}\)
Correct Answer: \(0.340\,\text{g}\)
Explanation: \( \textbf{Carbon mass from \( \mathrm{CO_2} \):} \)
\[
m_{\mathrm{C}}=1.320\times\frac{12}{44}=0.360\,\text{g}
\]
\( \textbf{Hydrogen mass from \( \mathrm{H_2O} \):} \)
\[
m_{\mathrm{H}}=0.540\times\frac{2}{18}=0.0600\,\text{g}
\]
\( \textbf{Nitrogen mass from percentage:} \)
\[
m_{\mathrm{N}}=\frac{15.56}{100}\times0.900=0.140\,\text{g}
\]
\( \textbf{Sum of known element masses:} \)
\[
0.360+0.0600+0.140=0.560\,\text{g}
\]
\( \textbf{Oxygen mass by difference:} \)
\[
m_{\mathrm{O}}=0.900-0.560=0.340\,\text{g}
\]
\( \textbf{Final answer:} \) The sample contains \(0.340\,\text{g}\) oxygen.
The difference method is valid here because the compound is stated to contain only carbon, hydrogen, nitrogen, and oxygen.
503. An elemental analysis gives \(0.360\,\text{g}\) carbon, \(0.0600\,\text{g}\) hydrogen, \(0.140\,\text{g}\) nitrogen, and \(0.320\,\text{g}\) oxygen. What empirical formula follows?
ⓐ. \( \mathrm{C_6H_{12}N_2O_4} \)
ⓑ. \( \mathrm{C_3H_6NO_2} \)
ⓒ. \( \mathrm{C_3H_6NO} \)
ⓓ. \( \mathrm{C_2H_4NO_2} \)
Correct Answer: \( \mathrm{C_3H_6NO_2} \)
Explanation: \( \textbf{Moles of carbon:} \)
\[
\frac{0.360}{12}=0.0300
\]
\( \textbf{Moles of hydrogen:} \)
\[
\frac{0.0600}{1}=0.0600
\]
\( \textbf{Moles of nitrogen:} \)
\[
\frac{0.140}{14}=0.0100
\]
\( \textbf{Moles of oxygen:} \)
\[
\frac{0.320}{16}=0.0200
\]
\( \textbf{Divide by the smallest value:} \)
\[
0.0300:0.0600:0.0100:0.0200=3:6:1:2
\]
\( \textbf{Final answer:} \) The empirical formula is \( \mathrm{C_3H_6NO_2} \).
The empirical formula is based on the simplest whole-number mole ratio of the elements.
504. An empirical-formula dataset gives \(0.360\,\text{g}\) carbon, \(0.0600\,\text{g}\) hydrogen, \(0.140\,\text{g}\) nitrogen, and \(0.320\,\text{g}\) oxygen in a compound sample. If the molar mass is \(88\,\text{g mol}^{-1}\), what is the molecular formula?
ⓐ. \( \mathrm{C_6H_{12}N_2O_4} \)
ⓑ. \( \mathrm{C_3H_6NO_2} \)
ⓒ. \( \mathrm{C_2H_6NO_3} \)
ⓓ. \( \mathrm{C_3H_8NO_2} \)
Correct Answer: \( \mathrm{C_3H_6NO_2} \)
Explanation: \( \textbf{Moles of carbon:} \)
\[
\frac{0.360}{12}=0.0300
\]
\( \textbf{Moles of hydrogen:} \)
\[
\frac{0.0600}{1}=0.0600
\]
\( \textbf{Moles of nitrogen:} \)
\[
\frac{0.140}{14}=0.0100
\]
\( \textbf{Moles of oxygen:} \)
\[
\frac{0.320}{16}=0.0200
\]
\( \textbf{Simplest ratio after dividing by \(0.0100\):} \)
\[
3:6:1:2
\]
\( \textbf{Empirical formula:} \)
\[
\mathrm{C_3H_6NO_2}
\]
\( \textbf{Empirical formula mass:} \)
\[
3(12)+6(1)+14+2(16)=36+6+14+32=88\,\text{g mol}^{-1}
\]
\( \textbf{Molar mass comparison:} \)
\[
\frac{88}{88}=1
\]
\( \textbf{Final answer:} \) The molecular formula is \( \mathrm{C_3H_6NO_2} \).
The empirical formula is already the molecular formula because its formula mass equals the given molar mass.
505. A set of analytical observations is given for an unknown organic compound:
| Observation | Inference |
|---|
| P. \( \mathrm{CO_2} \) turns lime water milky after combustion | carbon present |
| Q. anhydrous \( \mathrm{CuSO_4} \) turns blue after combustion | hydrogen present |
| R. Prussian blue from sodium fusion extract | nitrogen present |
| S. white precipitate with \( \mathrm{AgNO_3} \) after \( \mathrm{HNO_3} \) treatment, soluble in \( \mathrm{NH_4OH} \) | iodine present |
Which inference is unsuitable?
ⓐ. S
ⓑ. P
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: Lime water turns milky due to \( \mathrm{CO_2} \), so it supports the presence of carbon. Anhydrous \( \mathrm{CuSO_4} \) turns blue in the presence of water, which is produced from hydrogen during combustion. Prussian blue in Lassaigne’s test supports the presence of nitrogen. A white silver halide precipitate soluble in \( \mathrm{NH_4OH} \) indicates \( \mathrm{AgCl} \), so chlorine is indicated, not iodine. Iodine would give yellow \( \mathrm{AgI} \), which is insoluble in \( \mathrm{NH_4OH} \).
506. A compound is tested by several methods. Combustion shows \( \mathrm{C} \) and \( \mathrm{H} \), sodium fusion extract gives no nitrogen or sulfur test, and the halogen test gives pale yellow \( \mathrm{AgBr} \). Carius analysis gives \(40.0\%\) bromine. What extra information is still needed to write the molecular formula?
ⓐ. colour of the sodium fusion extract only
ⓑ. molar mass or vapour density
ⓒ. whether \( \mathrm{AgBr} \) is heavier than \( \mathrm{AgCl} \)
ⓓ. the shape of the test tube
Correct Answer: molar mass or vapour density
Explanation: Elemental tests and percentage analysis show which elements are present and how much of each is present by mass. From percentages, an empirical formula can be calculated. To convert the empirical formula into a molecular formula, the actual molar mass is needed. Vapour density can also provide molar mass through \(M=2\times\text{vapour density}\). Without molar mass information, only the simplest ratio can usually be assigned.
507. A compound has empirical formula \( \mathrm{C_2H_3Br} \). Its molar mass is \(321\,\text{g mol}^{-1}\). What molecular formula follows?
ⓐ. \( \mathrm{C_2H_3Br} \)
ⓑ. \( \mathrm{C_4H_6Br_2} \)
ⓒ. \( \mathrm{C_6H_9Br_3} \)
ⓓ. \( \mathrm{C_8H_{12}Br_4} \)
Correct Answer: \( \mathrm{C_6H_9Br_3} \)
Explanation: \( \textbf{Empirical formula mass:} \)
\[
2(12)+3(1)+80=24+3+80=107\,\text{g mol}^{-1}
\]
\( \textbf{Molar-mass multiplier:} \)
\[
\frac{321}{107}=3
\]
\( \textbf{Multiply each empirical subscript by \(3\):} \)
\[
\mathrm{C_2H_3Br}\times3=\mathrm{C_6H_9Br_3}
\]
\( \textbf{Final answer:} \) The molecular formula is \( \mathrm{C_6H_9Br_3} \).
A molecular formula must be an exact whole-number multiple of the empirical formula.
508. A correctly measured compound has empirical formula \( \mathrm{C_2H_3Br} \) and molar mass \(214\,\text{g mol}^{-1}\). What is the molecular formula?
ⓐ. \( \mathrm{C_2H_3Br} \)
ⓑ. \( \mathrm{C_6H_9Br_3} \)
ⓒ. \( \mathrm{C_8H_{12}Br_4} \)
ⓓ. \( \mathrm{C_4H_6Br_2} \)
Correct Answer: \( \mathrm{C_4H_6Br_2} \)
Explanation: \( \textbf{Empirical formula:} \)
\[
\mathrm{C_2H_3Br}
\]
\( \textbf{Empirical formula mass:} \)
\[
2(12)+3(1)+80=107\,\text{g mol}^{-1}
\]
\( \textbf{Molar mass:} \)
\[
214\,\text{g mol}^{-1}
\]
\( \textbf{Multiplier:} \)
\[
\frac{214}{107}=2
\]
\( \textbf{Multiply empirical subscripts by \(2\):} \)
\[
\mathrm{C_2H_3Br}\times2=\mathrm{C_4H_6Br_2}
\]
\( \textbf{Final answer:} \) The molecular formula is \( \mathrm{C_4H_6Br_2} \).
The empirical formula is a simplest unit, and the molar mass tells how many such units are present in one molecule.
509. A final analytical report says: “Carbon and hydrogen were detected by combustion, nitrogen by Prussian blue, chlorine by \( \mathrm{AgCl} \), and sulfur by \( \mathrm{BaSO_4} \) after oxidation.” Which statement best evaluates the report?
ⓐ. the report is impossible because one compound cannot contain more than two elements
ⓑ. the report combines qualitative detection and quantitative gravimetric evidence
ⓒ. the report proves the compound is an alkane
ⓓ. the report uses chromatography to identify every element
Correct Answer: the report combines qualitative detection and quantitative gravimetric evidence
Explanation: Combustion observations can indicate carbon and hydrogen qualitatively. The Prussian blue test detects nitrogen in sodium fusion extract. Chlorine can be estimated gravimetrically as \( \mathrm{AgCl} \), and sulfur can be estimated as \( \mathrm{BaSO_4} \). These methods can be combined to analyse different elements in one compound. The presence of heteroatoms such as nitrogen, chlorine, and sulfur means the compound is not a simple alkane.
510. A sodium fusion extract gives a violet colour with sodium nitroprusside and, after separate treatment with dilute \( \mathrm{HNO_3} \) followed by \( \mathrm{AgNO_3} \), gives a pale yellow precipitate partly soluble in \( \mathrm{NH_4OH} \). What elements are indicated?
ⓐ. sulfur and bromine
ⓑ. nitrogen and chlorine
ⓒ. sulfur and iodine
ⓓ. nitrogen and fluorine
Correct Answer: sulfur and bromine
Explanation: Sodium nitroprusside gives a violet colour with sulfide ion, so the sodium fusion extract indicates sulfur. The halogen test must be done after boiling with dilute \( \mathrm{HNO_3} \) so that interfering \( \mathrm{CN^-} \) and \( \mathrm{S^{2-}} \) ions are removed. A pale yellow precipitate with \( \mathrm{AgNO_3} \) that is partly soluble in ammonium hydroxide is characteristic of \( \mathrm{AgBr} \). Chloride would give white \( \mathrm{AgCl} \), while iodide would give yellow \( \mathrm{AgI} \) insoluble in ammonium hydroxide. The observations therefore support sulfur and bromine together.
511. A student writes the following inference table from qualitative organic analysis.
| Observation | Student inference |
|---|
| P. Prussian blue colour in Lassaigne’s nitrogen test | nitrogen present |
| Q. Violet colour with sodium nitroprusside | sulfur present |
| R. Yellow \( \mathrm{AgI} \) precipitate insoluble in \( \mathrm{NH_4OH} \) | iodine present |
| S. White \( \mathrm{AgCl} \) precipitate soluble in \( \mathrm{NH_4OH} \) | bromine present |
Which inference is unsuitable?
ⓐ. S
ⓑ. P
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: Prussian blue colour is the usual positive indication for nitrogen in the sodium fusion extract. A violet colour with sodium nitroprusside indicates sulfur because sulfide ion reacts with the reagent. Yellow \( \mathrm{AgI} \) insoluble in ammonium hydroxide indicates iodine. A white silver halide precipitate soluble in ammonium hydroxide is \( \mathrm{AgCl} \), not \( \mathrm{AgBr} \). Bromide gives pale yellow \( \mathrm{AgBr} \), whose solubility in ammonium hydroxide is much less than that of \( \mathrm{AgCl} \).
512. In a Carius estimation, \(0.420\,\text{g}\) of an organic compound containing chlorine gives \(0.765\,\text{g}\) of \( \mathrm{AgCl} \). If another analysis gives \(40.0\%\) carbon and \(5.0\%\) hydrogen, what is the percentage of oxygen assuming only \( \mathrm{C} \), \( \mathrm{H} \), \( \mathrm{O} \), and \( \mathrm{Cl} \) are present? Use \( \mathrm{AgCl}=143.5 \) and \( \mathrm{Cl}=35.5 \).
ⓐ. \(5.0\%\)
ⓑ. \(20.0\%\)
ⓒ. \(10.0\%\)
ⓓ. \(30.0\%\)
Correct Answer: \(10.0\%\)
Explanation: \( \textbf{Mass of chlorine in \( \mathrm{AgCl} \):} \)
\[
m_{\mathrm{Cl}}=0.765\times\frac{35.5}{143.5}=0.189\,\text{g}
\]
\( \textbf{Percentage of chlorine in sample:} \)
\[
\%\mathrm{Cl}=\frac{0.189}{0.420}\times100\approx45.0\%
\]
\( \textbf{Given percentages:} \)
\[
\%\mathrm{C}=40.0,\quad \%\mathrm{H}=5.0
\]
\( \textbf{Oxygen by difference:} \)
\[
\%\mathrm{O}=100-(40.0+5.0+45.0)=10.0\%
\]
\( \textbf{Final answer:} \) The oxygen percentage is \(10.0\%\).
Chlorine is first calculated from the silver chloride precipitate, and oxygen is then obtained by difference.
513. In a second Carius estimation, \(0.420\,\text{g}\) of compound gives \(0.764\,\text{g}\) of \( \mathrm{AgCl} \). If the same compound contains \(40.0\%\) carbon and \(5.0\%\) hydrogen, what is the oxygen percentage assuming only \( \mathrm{C} \), \( \mathrm{H} \), \( \mathrm{O} \), and \( \mathrm{Cl} \) are present? Use \( \mathrm{AgCl}=143.5 \) and \( \mathrm{Cl}=35.5 \).
ⓐ. \(5.0\%\)
ⓑ. \(10.0\%\)
ⓒ. \(15.0\%\)
ⓓ. \(20.0\%\)
Correct Answer: \(10.0\%\)
Explanation: \( \textbf{Chlorine mass from \( \mathrm{AgCl} \):} \)
\[
m_{\mathrm{Cl}}=0.764\times\frac{35.5}{143.5}=0.189\,\text{g}
\]
\( \textbf{Percentage chlorine:} \)
\[
\%\mathrm{Cl}=\frac{0.189}{0.420}\times100=45.0\%
\]
\( \textbf{Known percentages:} \)
\[
\%\mathrm{C}=40.0,\quad \%\mathrm{H}=5.0,\quad \%\mathrm{Cl}=45.0
\]
\( \textbf{Oxygen by difference:} \)
\[
\%\mathrm{O}=100-(40.0+5.0+45.0)
\]
\[
\%\mathrm{O}=100-90.0=10.0\%
\]
\( \textbf{Final answer:} \) The oxygen percentage is \(10.0\%\).
The calculation is valid because the compound is stated to contain only carbon, hydrogen, oxygen, and chlorine.
514. A compound containing only \( \mathrm{C} \), \( \mathrm{H} \), \( \mathrm{O} \), and \( \mathrm{Cl} \) has \(36.0\%\) \( \mathrm{C} \), \(5.0\%\) \( \mathrm{H} \), \(16.0\%\) \( \mathrm{O} \), and \(43.0\%\) \( \mathrm{Cl} \). Which empirical formula is obtained? Use \( \mathrm{C}=12 \), \( \mathrm{H}=1 \), \( \mathrm{O}=16 \), and \( \mathrm{Cl}=35.5 \).
ⓐ. \( \mathrm{C_3H_5OCl} \)
ⓑ. \( \mathrm{C_2H_4OCl_2} \)
ⓒ. \( \mathrm{C_4H_6OCl} \)
ⓓ. \( \mathrm{C_3H_6O_2Cl} \)
Correct Answer: \( \mathrm{C_3H_5OCl} \)
Explanation: \( \textbf{Assume \(100\,\text{g}\) compound:} \)
\[
\mathrm{C}=36.0\,\text{g},\quad \mathrm{H}=5.0\,\text{g},\quad \mathrm{O}=16.0\,\text{g},\quad \mathrm{Cl}=43.0\,\text{g}
\]
\( \textbf{Moles of carbon:} \)
\[
\frac{36.0}{12}=3.00
\]
\( \textbf{Moles of hydrogen:} \)
\[
\frac{5.0}{1}=5.00
\]
\( \textbf{Moles of oxygen:} \)
\[
\frac{16.0}{16}=1.00
\]
\( \textbf{Moles of chlorine:} \)
\[
\frac{43.0}{35.5}\approx1.21\approx1.00
\]
\( \textbf{Simplest whole-number ratio:} \)
\[
\mathrm{C:H:O:Cl}=3:5:1:1
\]
\( \textbf{Empirical formula:} \)
\[
\mathrm{C_3H_5OCl}
\]
\( \textbf{Final answer:} \) The empirical formula is \( \mathrm{C_3H_5OCl} \).
The chlorine percentage is rounded to the nearest whole percent, so \(43.0\%\) is close to the value expected for one chlorine atom in this empirical formula.
515. In an elemental-analysis problem, the mole ratio after division by the smallest mole value is \( \mathrm{C:H:Cl}=1.00:1.50:0.50 \). Which empirical formula should be written?
ⓐ. \( \mathrm{CH_{1.5}Cl_{0.5}} \)
ⓑ. \( \mathrm{CH_3Cl} \)
ⓒ. \( \mathrm{C_2H_2Cl} \)
ⓓ. \( \mathrm{C_2H_3Cl} \)
Correct Answer: \( \mathrm{C_2H_3Cl} \)
Explanation: Empirical formulas must use the simplest whole-number ratio of atoms. The ratio \(1.00:1.50:0.50\) contains fractional terms. Multiplying every term by \(2\) gives \(2.00:3.00:1.00\). Therefore, the empirical formula is \( \mathrm{C_2H_3Cl} \). Only one part of the ratio should not be multiplied alone, because all subscripts must preserve the relative atom ratio.
516. A compound has empirical formula \( \mathrm{C_2H_3Cl} \). Its vapour density is \(62.5\). What molecular formula is obtained? Use \( \mathrm{C}=12 \), \( \mathrm{H}=1 \), and \( \mathrm{Cl}=35.5 \).
ⓐ. \( \mathrm{C_4H_6Cl_2} \)
ⓑ. \( \mathrm{C_2H_3Cl} \)
ⓒ. \( \mathrm{C_6H_9Cl_3} \)
ⓓ. \( \mathrm{C_8H_{12}Cl_4} \)
Correct Answer: \( \mathrm{C_4H_6Cl_2} \)
Explanation: \( \textbf{Empirical formula:} \)
\[
\mathrm{C_2H_3Cl}
\]
\( \textbf{Empirical formula mass:} \)
\[
2(12)+3(1)+35.5=24+3+35.5=62.5
\]
\( \textbf{Molar mass from vapour density:} \)
\[
M=2\times62.5=125\,\text{g mol}^{-1}
\]
\( \textbf{Multiplier:} \)
\[
\frac{125}{62.5}=2
\]
\( \textbf{Molecular formula:} \)
\[
\mathrm{C_4H_6Cl_2}
\]
\( \textbf{Final answer:} \) The molecular formula is \( \mathrm{C_4H_6Cl_2} \).
Vapour density gives molar mass, and the molar mass decides how many empirical units are present in one molecule.
517. An organic compound is analysed as follows: carbon and hydrogen are found by combustion, nitrogen by Dumas method, and chlorine by Carius method. Oxygen is then calculated by difference. Which condition is essential for the oxygen-by-difference step?
ⓐ. the compound must contain no carbon
ⓑ. the compound must give no \( \mathrm{CO_2} \)
ⓒ. the Carius precipitate must be ignored
ⓓ. all other elements must be accounted for
Correct Answer: all other elements must be accounted for
Explanation: Oxygen by difference means subtracting the percentages of all other elements from \(100\%\). This works only if every other element present has been detected and estimated. If sulfur, phosphorus, halogen, or nitrogen is present but not included, the difference will be wrongly assigned to oxygen. Carbon and hydrogen values must also be correctly calculated from \( \mathrm{CO_2} \) and \( \mathrm{H_2O} \). The difference method is therefore a bookkeeping step that depends on complete elemental accounting.
518. A compound contains \( \mathrm{C} \), \( \mathrm{H} \), \( \mathrm{N} \), and \( \mathrm{Cl} \), but no oxygen. A student subtracts only carbon and hydrogen from \(100\%\) and calls the remaining mass oxygen. What is the main error?
ⓐ. oxygen must always be present in organic compounds
ⓑ. carbon cannot be estimated from \( \mathrm{CO_2} \)
ⓒ. chlorine cannot be estimated gravimetrically
ⓓ. nitrogen and chlorine were ignored
Correct Answer: nitrogen and chlorine were ignored
Explanation: The difference method assigns the unmeasured remainder to an element only when all other elements present have already been included. Here the compound contains nitrogen and chlorine. If the student subtracts only carbon and hydrogen, the remaining percentage includes nitrogen and chlorine, not oxygen. Since the compound is stated to contain no oxygen, the conclusion is clearly wrong. Elemental analysis requires first knowing which elements are present and then accounting for each one properly.
519. A compound gives \(0.440\,\text{g}\) \( \mathrm{CO_2} \), \(0.180\,\text{g}\) \( \mathrm{H_2O} \), and \(0.1435\,\text{g}\) \( \mathrm{AgCl} \) from separate suitable samples. Which calculation correctly gives the mass of chlorine from \( \mathrm{AgCl} \)?
ⓐ. \(0.1435\times\frac{108}{143.5}\)
ⓑ. \(0.1435\times\frac{143.5}{35.5}\)
ⓒ. \(0.1435\times\frac{35.5}{143.5}\)
ⓓ. \(0.1435+35.5\)
Correct Answer: \(0.1435\times\frac{35.5}{143.5}\)
Explanation: Silver chloride contains both silver and chlorine. The fraction of chlorine in \( \mathrm{AgCl} \) is the atomic mass of chlorine divided by the molar mass of silver chloride. Since \(M(\mathrm{AgCl})=108+35.5=143.5\), the chlorine fraction is \( \frac{35.5}{143.5} \). Multiplying the precipitate mass by this fraction gives the chlorine mass. Using the silver fraction would calculate silver mass, not chlorine mass.
520. A graph is drawn for an empirical-formula calculation. The horizontal axis shows the calculation stage: mass of element, moles of element, divided ratio, and whole-number ratio. The graph shows a sharp change between mass and moles for chlorine compared with hydrogen. What is the best reason?
ⓐ. chlorine cannot be part of an empirical formula
ⓑ. hydrogen is always absent when chlorine is present
ⓒ. chlorine's larger atomic mass gives fewer moles
ⓓ. mole ratios are obtained by adding atomic masses to percentages
Correct Answer: chlorine's larger atomic mass gives fewer moles
Explanation: Empirical formulas are based on mole ratios, not direct mass ratios. Chlorine has atomic mass about \(35.5\), while hydrogen has atomic mass \(1\). Therefore, a given mass of chlorine corresponds to far fewer moles than the same mass of hydrogen. This is why the conversion from mass to moles can change the relative scale strongly. The graph highlights the need to divide each element mass by its own atomic mass before writing subscripts.
