Stoichiometry and Empirical Formulas

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Questions and Answers

The empirical formula of copper oxide is Cu2O.

True (A)

If a student burns 9 g of magnesium in excess oxygen, 25 g of magnesium oxide will be produced.

False (B)

To produce magnesium oxide, the reaction requires 2 moles of magnesium for every mole of oxygen.

True (A)

56 liters of carbon dioxide is produced from burning 30 g of carbon in excess oxygen.

<p>True (A)</p> Signup and view all the answers

The mass of oxygen in copper oxide calculated was 0.6 g.

<p>True (A)</p> Signup and view all the answers

The empirical formula of glucose is C6H12O6.

<p>False (B)</p> Signup and view all the answers

The percentage composition of CaO by mass is 71.43% calcium and 28.57% oxygen.

<p>True (A)</p> Signup and view all the answers

To find the relative molecular mass of CaO, you add the atomic masses of calcium and oxygen.

<p>True (A)</p> Signup and view all the answers

In the formula of glucose, the value of n for converting the empirical formula to the molecular formula is 5.

<p>False (B)</p> Signup and view all the answers

Stoichiometry is only concerned with the amount of products in a chemical reaction.

<p>False (B)</p> Signup and view all the answers

The empirical formula is identical to the molecular formula for all compounds.

<p>False (B)</p> Signup and view all the answers

The relative molecular mass of glucose is 150.

<p>False (B)</p> Signup and view all the answers

The molecular formula can be derived from the empirical formula using a whole number multiplier.

<p>True (A)</p> Signup and view all the answers

Flashcards

Stoichiometry

The study of the quantitative relationships between reactants and products in a chemical reaction.

Percentage composition

The percentage by mass of each element in a compound.

Empirical formula

The simplest whole-number ratio of atoms in a compound.

Molecular formula

The actual number and type of atoms in a molecule.

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Relative Molecular Mass

The sum of the atomic masses of all atoms in a molecule.

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Empirical Formula Calculation (from % composition)

Determining the empirical formula from the percentages of each element.

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Molecular formula calculation

Calculating the molecular formula from the empirical formula and relative molecular mass.

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Empirical Formula (from reaction data)

Determining the empirical formula from the masses of reactants and products in a chemical reaction

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Empirical Formula of Copper Oxide

The simplest whole-number ratio of copper and oxygen atoms in copper oxide.

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Calculating mass of magnesium oxide from mass of magnesium

Using balanced chemical equation 2Mg + O₂ → 2MgO , to determine the amount of magnesium oxide formed from a given mass of magnesium.

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Calculating volume of CO2 from mass of carbon

Determining the volume of CO2 produced when a specific mass of carbon is burned in excess oxygen, using the balanced chemical equation C + O₂ → CO₂.

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Moles

A unit of measurement representing a specific amount of a substance.

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Balanced Chemical Equation

A chemical equation where the number of atoms of each element is equal on both sides of the equation.

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Study Notes

Stoichiometry

  • Stoichiometry is the study of the relationships between reactants and products in a chemical reaction.
  • Chemists use stoichiometry to determine the amounts of reactants needed to produce a certain amount of product.

Calculating Percentage Composition

  • To find the percentage composition of an element in a compound, calculate the relative molecular mass of the compound.
  • Divide the atomic mass of the element by the relative molecular mass of the compound, then multiply by 100%.

Empirical Formula

  • Molecular formula: Shows the actual number and type of atoms in a molecule.
  • Empirical formula: Shows the simplest whole number ratio of atoms in a compound.
  • Molecular formula = empirical formula x n (where n is a whole number)

Calculating Empirical Formula from Masses of Reactants and Products

  • An experiment was carried out to find the empirical (simplest) formula of a copper oxide.
  • 5.4 g of the oxide gave 4.8 g of copper.
  • Find the mass of oxygen in the oxide (5.4 g – 4.8 g = 0.6 g).
  • Calculate the number of moles of copper and oxygen.
  • Divide the number of moles of each element by the smallest number of moles to get the simplest whole number ratio. This will be the subscripts of the elements in the empirical formula.

Calculating Masses of Reactants or Products from Balanced Chemical Equations

  • The balanced equation for the reaction between magnesium and oxygen to form magnesium oxide is : 2Mg + O2 → 2MgO
  • If 9 g of magnesium is burned in excess oxygen, the mass of magnesium oxide formed can be calculated using the molar masses and the stoichiometric ratios from the balanced equation.

Calculating Volumes of Gases

  • The balanced equation for the reaction between carbon and oxygen to produce carbon dioxide is: C + O2 → CO2
  • If 30 g of carbon is burned in excess oxygen, the volume of carbon dioxide produced at standard temperature and pressure (STP) can be calculated by using the molar volume of a gas, which is 22.4 L/mol.

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