How do you exactly do gravimetric stoichiometry?

Gravimetric stoichiometry is the branch of stoichiometry that deals with using the mass of one substance to calculate the mass of another substance in a chemical reaction.

For example, in the reaction

##”A → B”##

gravimetric stoichiometry asks the questions, “What mass of ##”B”## can be formed from ##x## g of ##”A”##?” and “What mass of ##”A”## is required to form ##x## g of ##”B”##?”

You always use the same type of steps.

  • Write the balanced chemical equation.
  • Convert ##”mass of A” → “moles of A” → “moles of B” → “mass of B”##.

EXAMPLE

Calculate the mass of iron(III) oxide that can be produced by the reaction of 500 g of iron with excess oxygen.

Solution

Step 1. Write the balanced chemical equation.

##”4Fe” + “3O”_2 → “2Fe”_2″O”_3##

Step 2. Do the conversions.

##500 color(red)(cancel(color(black)(“g Fe”))) × (1color(red)(cancel(color(black)(“mol Fe”))))/(55.84 color(red)(cancel(color(black)(“g Fe”)))) × (2 color(red)(cancel(color(black)(“mol Fe”_2″O”_3))))/(4 color(red)(cancel(color(black)(“mol Fe”)))) × (“159.7 g Fe”_2″O”_3)/(1 color(red)(cancel(color(black)(“mol Fe”_2″O”_3)))) = “715 g Fe”_2″O”_3##

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How do you exactly do gravimetric stoichiometry?

Gravimetric stoichiometry is the branch of stoichiometry that deals with using the mass of one substance to calculate the mass of another substance in a chemical reaction.

For example, in the reaction

##”A → B”##

gravimetric stoichiometry asks the questions, “What mass of ##”B”## can be formed from ##x## g of ##”A”##?” and “What mass of ##”A”## is required to form ##x## g of ##”B”##?”

You always use the same type of steps.

  • Write the balanced chemical equation.
  • Convert ##”mass of A” → “moles of A” → “moles of B” → “mass of B”##.

EXAMPLE

Calculate the mass of iron(III) oxide that can be produced by the reaction of 500 g of iron with excess oxygen.

Solution

Step 1. Write the balanced chemical equation.

##”4Fe” + “3O”_2 → “2Fe”_2″O”_3##

Step 2. Do the conversions.

##500 color(red)(cancel(color(black)(“g Fe”))) × (1color(red)(cancel(color(black)(“mol Fe”))))/(55.84 color(red)(cancel(color(black)(“g Fe”)))) × (2 color(red)(cancel(color(black)(“mol Fe”_2″O”_3))))/(4 color(red)(cancel(color(black)(“mol Fe”)))) × (“159.7 g Fe”_2″O”_3)/(1 color(red)(cancel(color(black)(“mol Fe”_2″O”_3)))) = “715 g Fe”_2″O”_3##

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