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Calculate the relative formula mass of calcium carbonate, CaCO₃ - Edexcel - GCSE Chemistry - Question 4 - 2016 - Paper 1

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Calculate the relative formula mass of calcium carbonate, CaCO₃. (relative atomic masses: C = 12, O = 16, Ca = 40) Magnesium reacts with sulfur to form magnesium su... show full transcript

Worked Solution & Example Answer:Calculate the relative formula mass of calcium carbonate, CaCO₃ - Edexcel - GCSE Chemistry - Question 4 - 2016 - Paper 1

Step 1

Calculate the relative formula mass of calcium carbonate, CaCO₃.

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Answer

To calculate the relative formula mass of calcium carbonate (CaCO₃), we sum the relative atomic masses of each element in the compound:

  • Calcium (Ca): 40 g/mol
  • Carbon (C): 12 g/mol
  • Oxygen (O): 16 g/mol, and there are 3 oxygen atoms. Therefore:

extRelativeFormulaMass=40+12+(3imes16)=40+12+48=100extg/mol ext{Relative Formula Mass} = 40 + 12 + (3 imes 16) = 40 + 12 + 48 = 100 ext{ g/mol}

Thus, the relative formula mass of CaCO₃ is 100 g/mol.

Step 2

The ratio of magnesium atoms to sulfur atoms in magnesium sulfide is

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Answer

In magnesium sulfide (MgS), the molar mass of magnesium (Mg) is 24 g/mol and sulfur (S) is 32 g/mol. From the combination:

  • 24 g of magnesium contains 1 mole of Mg.
  • 32 g of sulfur contains 1 mole of S.

Therefore, the ratio of magnesium atoms to sulfur atoms is: extRatio=1:1 ext{Ratio} = 1 : 1

Step 3

Give the empirical formula of aluminium oxide.

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Answer

The simplest ratio of aluminium atoms to oxygen atoms in aluminium oxide is given as 2 : 3. Thus, the empirical formula of aluminium oxide can be represented as:

extEmpiricalFormula=extAl2extO3 ext{Empirical Formula} = ext{Al}_2 ext{O}_3

Step 4

What is the molecular formula of the hydrocarbon with the empirical formula CH₂?

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The molecular formula can be determined by examining the given empirical formula (CH₂). Hydrocarbons with CH₂ as the empirical formula could have molecular formulas such as C₂H₄ (ethene) or C₃H₆ (propene). Therefore, the molecular formula options are:

  • A (C₄H₈)
  • B (C₂H₄)
  • C (C₃H₆)
  • D (C₄H₈)

The correct answer is C₂H₄, which can be marked with a cross (🗹).

Step 5

Calculate the percentage yield of zinc oxide.

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Answer

The percentage yield is calculated using the formula:

ext{Percentage Yield} = rac{ ext{Actual Yield}}{ ext{Theoretical Yield}} imes 100

Given:

  • Actual Yield = 7.2 g
  • Theoretical Yield = 9.0 g

Thus: ext{Percentage Yield} = rac{7.2}{9.0} imes 100 = 80 ext{ %}

Step 6

Explain why the actual yield of a product in an experiment is usually lower than the theoretical yield.

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Answer

The actual yield of a product is often lower than the theoretical yield due to several factors:

  1. Incomplete Reactions: Not all reactants may have reacted, leading to less product formation.
  2. Side Reactions: Other unwanted chemical reactions may occur, producing by-products instead of the desired product.
  3. Loss of Product: Some product may be lost during the transfer or purification processes, which lowers the measured yield.

Step 7

Calculate the percentage by mass of zinc in zinc oxide, ZnO.

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Answer

To find the percentage by mass of zinc in zinc oxide (ZnO), we first need to determine the molar mass of ZnO:

  • Zinc (Zn): 65 g/mol
  • Oxygen (O): 16 g/mol

Total molar mass of ZnO: extMolarMassofZnO=65+16=81extg/mol ext{Molar Mass of ZnO} = 65 + 16 = 81 ext{ g/mol}

The percentage by mass of zinc is given by: ext{Percentage by Mass of Zn} = rac{ ext{Molar Mass of Zn}}{ ext{Molar Mass of ZnO}} imes 100 = rac{65}{81} imes 100 \\approx 80.25 ext{ %}

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