This question assesses the skill of elemental composition of pure substances. The chemical formula Fe₂O₃ indicates two iron and three oxygen atoms per formula unit. The molar mass is 2×55.8 + 3×16.0 = 159.6 g/mol, with iron contributing 111.6 g/mol. The proportion of iron is 111.6/159.6 ≈ 0.699, so in a 5.00 g sample, the mass of iron is 0.699 × 5.00 ≈ 3.50 g. A tempting distractor is 2.00 g, which could result from incorrectly dividing the sample mass by the number of iron atoms without using molar masses. To solve similar problems, use the formula to count atoms, then convert to mass contribution via the ratio of elemental to total molar mass multiplied by sample mass.

This question assesses the skill of elemental composition of pure substances. The chemical formula K₂Cr₂O₇ shows two potassium, two chromium, and seven oxygen atoms per formula unit. While molar masses would determine mass proportions, here the atom ratio is directly from the formula, giving K:O as 2:7 without needing masses for count-based composition. This ratio reflects the relative number of each atom type in the compound. A tempting distractor is 2:1, which might come from miscounting oxygen atoms or ignoring the subscript on O. To solve similar problems, use the formula to count atoms, then convert to mass contribution if needed, or directly use counts for ratios.

This question assesses the skill of elemental composition of pure substances. The chemical formula (NH4)2SO4 indicates two nitrogen, eight hydrogen, one sulfur, and four oxygen atoms per formula unit. To find the mass proportion, calculate the molar mass using atomic masses: 14.0 g/mol for each N, 1.0 g/mol for each H, 32.1 g/mol for S, and 16.0 g/mol for each O, totaling 132.1 g/mol, with nitrogen contributing 28.0 g/mol. For a 13.2 g sample, the mass of nitrogen is (28.0 / 132.1) × 13.2 ≈ 2.80 g. A common error is counting only one NH4 group, leading to 1.40 g, which is incorrect because the formula has two NH4 groups. Always use the formula to count the number of each type of atom, then compute their total mass contribution using molar masses, and find the proportion.

This question assesses the skill of elemental composition of pure substances. The chemical formula NaHCO₃ shows one sodium, one hydrogen, one carbon, and three oxygen atoms. The molar mass is 23.0 + 1.0 + 12.0 + 3×16.0 = 84.0 g/mol, with oxygen contributing 48.0 g/mol. The proportion of oxygen is 48.0/84.0 ≈ 0.571, so in a 12.5 g sample, the mass of oxygen is 0.571 × 12.5 ≈ 7.14 g. A tempting distractor is 4.0 g, which might arise from using only one oxygen atom's mass instead of three in the calculation. To solve similar problems, use the formula to count atoms, then convert to mass contribution by determining the elemental mass fraction and applying it to the sample mass.

This question assesses the skill of elemental composition of pure substances. The chemical formula C₆H₁₂O₆ indicates six carbon, twelve hydrogen, and six oxygen atoms per molecule. The molar mass is 6×12.0 + 12×1.0 + 6×16.0 = 180.0 g/mol, with carbon contributing 72.0 g/mol. The proportion of carbon by mass is 72.0/180.0 = 0.40, so in an 18.0 g sample, the mass of carbon is 0.40 × 18.0 = 7.2 g. A tempting distractor is 12.0 g, which could result from mistakenly using only the atomic mass of carbon without accounting for the number of atoms or the total molar mass. To solve similar problems, use the formula to count atoms, then convert to mass contribution by finding the elemental fraction of the molar mass and multiplying by the sample mass.

This question assesses the skill of elemental composition of pure substances. The chemical formula CO₂ indicates one carbon and two oxygen atoms per molecule. The molar mass is 12.0 + 2×16.0 = 44.0 g/mol, with oxygen contributing 32.0 g/mol. The proportion of oxygen is 32.0/44.0 ≈ 0.727, so in a 30.0 g sample, the mass of oxygen is 0.727 × 30.0 ≈ 21.8 g. A tempting distractor is 16.0 g, which might come from using only one oxygen atom's mass without accounting for two or the total molar mass. To solve similar problems, use the formula to count atoms, then convert to mass contribution by calculating the elemental fraction of the molar mass and multiplying by the sample mass.

This question assesses the skill of elemental composition of pure substances. The chemical formula NH₄Cl shows one nitrogen, four hydrogen, and one chlorine atom per formula unit. Using atomic masses, the molar mass is 14.0 + 4×1.0 + 35.5 = 53.5 g/mol, with nitrogen contributing 14.0 g/mol. The mass percent of nitrogen is (14.0/53.5) × 100 ≈ 26.2%, determined by the ratio of nitrogen's mass to the total molar mass. A tempting distractor is 31.0%, which might come from incorrectly using chlorine's mass instead of nitrogen's in the percentage calculation. To solve similar problems, use the formula to count atoms, then convert to mass contribution by computing the percentage based on elemental and total molar masses.

This question assesses the skill of elemental composition of pure substances. The chemical formula Fe2O3 indicates two iron atoms and three oxygen atoms per formula unit. To find the mass proportion, calculate the molar mass using atomic masses: 55.8 g/mol for each Fe and 16.0 g/mol for each O, totaling 159.6 g/mol, with iron contributing 111.6 g/mol. For a 16.0 g sample, the mass of iron is (111.6 / 159.6) × 16.0 ≈ 11.2 g. A common error is calculating for one Fe atom, yielding ≈5.60 g, which is incorrect because the formula has two Fe atoms. Always use the formula to count the number of each type of atom, then compute their total mass contribution using molar masses, and find the proportion.

This question assesses the skill of elemental composition of pure substances. The chemical formula H2O2 indicates two hydrogen atoms and two oxygen atoms per formula unit. To find the mass proportion, calculate the molar mass using atomic masses: 1.0 g/mol for each H and 16.0 g/mol for each O, totaling 34.0 g/mol, with hydrogen contributing 2.0 g/mol. The mass percent of hydrogen is (2.0 / 34.0) × 100 ≈ 5.88%. A common error is calculating percent oxygen instead, yielding 94.1%, which is incorrect because the question asks for hydrogen. Always use the formula to count the number of each type of atom, then compute their total mass contribution using molar masses, and find the percentage.

This question assesses the skill of elemental composition of pure substances. The chemical formula $\mathrm{Li_2CO_3}$ indicates two lithium, one carbon, and three oxygen atoms per formula unit. To find the mass proportion, calculate the molar mass using atomic masses: 6.94 g/mol for each Li, 12.0 g/mol for C, and 16.0 g/mol for each O, totaling 73.88 g/mol, with oxygen contributing 48.0 g/mol. For a 7.40 g sample, the mass of oxygen is $(48.0 / 73.88) \times 7.40 \approx 4.80 \mathrm{g}$. A common error is calculating for two O atoms, yielding $\approx 3.20 \mathrm{g}$, which is incorrect because the formula has three O atoms. Always use the formula to count the number of each type of atom, then compute their total mass contribution using molar masses, and find the proportion.