l,{"backHref":"/gcse/chemistry/edexcel/questions","basePath":"/gcse/chemistry/edexcel/calculations-involving-masses/questions","formulaUrl":null,"initialIndex":49,"initialQuestionWindow":[{"id":"2c2a12be-f86f-45ff-9da2-2747682d554d","revisionId":"64ebc650-74f2-456f-87c3-11b7fdd88d78","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"7611fd28-489d-4fb6-b263-5d9ded472ba3","label":null,"content":"Two oxides of copper, **X** and **Y**, were analysed.\n\nThe empirical formula of oxide **X** was found to be $\\text{CuO}$.\n\nThe results of the analysis of oxide **Y** showed it contained 1.27 g of copper combined with 0.16 g of oxygen.\n\nShow, by calculation, that the two oxides had different empirical formulae.\n(relative atomic masses: $\\text{O} = 16$, $\\text{Cu} = 63.5$)","markscheme":"- Calculates the amount of copper in moles: $\\text{moles of Cu} = \\frac{1.27}{63.5} = 0.02\\text{ mol}$ M1\n- Calculates the amount of oxygen in moles: $\\text{moles of O} = \\frac{0.16}{16} = 0.01\\text{ mol}$ M1\n- Determines the simplest ratio of $\\text{Cu} : \\text{O} = 2 : 1$, giving the empirical formula of Y as $\\text{Cu}_2\\text{O}$ and concluding that this is different from oxide X ($\\text{CuO}$) A1","marks":3,"order":0,"markingRubric":null,"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"2c6c7695-6216-408b-b57e-ac4e57f82cf6","revisionId":"931ce183-c936-4462-96a5-bf8b3f151622","questionNumber":null,"questionType":"LA","specification":"","difficulty":6,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"ba637c98-fb12-46ce-88f0-4361ac57a84b","label":null,"content":"In another experiment, $324\\text{ mg}$ of silver is deposited at the negative electrode.\n\nCalculate the number of silver atoms in $324\\text{ mg}$ of silver.\n\nGive your answer to $3$ significant figures.\n\n($\\text{relative atomic mass } A_r(\\text{Ag}) = 108$; Avogadro constant = $6.02 \\times 10^{23}$ per mole)","markscheme":"- Converts $324\\text{ mg}$ to $0.324\\text{ g}$ B1\n- Calculates the number of moles of silver: $\\text{moles} = \\frac{0.324}{108} = 0.003\\text{ mol}$ M1\n- Calculates the number of silver atoms and rounds to $3$ significant figures: $0.003 \\times 6.02 \\times 10^{23} = 1.81 \\times 10^{21}$ A1","marks":3,"order":0,"markingRubric":null,"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"31a471a8-fab5-4c8e-b11a-62d76fe20231","revisionId":"d3f2d298-3e0e-4980-8336-3fb97ae73787","questionNumber":null,"questionType":"LA","specification":"","difficulty":5,"difficultyBand":"medium","totalMarks":3,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"66eccc04-be53-4eed-8dfd-d93f7f9aece3","label":null,"content":"Ammonium phosphate, $(\\text{NH}_4)_3\\text{PO}_4$, is a highly concentrated salt commonly used in agricultural fertilizers to promote healthy plant growth.\n\nCalculate the percentage by mass of nitrogen in ammonium phosphate.\n\n(Relative atomic masses: $\\text{H} = 1.0$, $\\text{N} = 14.0$, $\\text{O} = 16.0$, $\\text{P} = 31.0$)","markscheme":"- Calculates the relative formula mass ($M_r$) of ammonium phosphate: $M_r = (3 \\times 14.0) + (12 \\times 1.0) + 31.0 + (4 \\times 16.0) = 149.0$ M1\n- Sets up the percentage by mass calculation: $\\frac{42.0}{149.0} \\times 100$ M1\n- Calculates the final percentage: $28.2\\%$ (accept $28\\%$ or $28.19\\%$) A1","marks":3,"order":0,"markingRubric":null,"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null}],"subjectCode":"gcse_chemistry","summaries":[{"id":"0022739c-a336-4f02-b237-23301e5ab290","difficultyBand":"medium","calculatorUse":null},{"id":"00a695e1-8f94-48c2-9ede-d32dde228113","difficultyBand":"medium","calculatorUse":null},{"id":"03c04d6b

Calculations involving masses

Question 25

Calculations involving masses Questions

Calculations involving masses

Question 25

Calculations involving masses Questions