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Experiment 2: stoichiometry: laws to moles to molarity


  • To determine the mass of a product of a chemical reaction
  • To make a solution of assigned molarity–your accuracy will be tested by your TA by titration!


  • Pre-lab (10%)
  • Lab Report (80%)
  • TA points (10%)

Before coming to lab..

  • Read the lab instructions
  • Complete the pre-lab, due at the beginning of the lab


The word stoichiometry derives from two Greek words: stoicheion (meaning "element") and metron (meaning "measure"). Stoichiometry deals with calculations about the masses (sometimes volumes) of reactants and products involved in a chemical reaction. Consequently, it is a very mathematical part of chemistry.

In the first part of this lab, sodium bicarbonate is reacted with an excess of hydrochloric acid.

NaHCO 3 ( s ) + HCl ( aq ) NaCl ( aq ) + CO 2 ( g ) + H 2 O size 12{"NaHCO" rSub { size 8{3} } \( s \) +"HCl" \( "aq" \) rightarrow "NaCl" \( "aq" \) +"CO" rSub { size 8{2} } \( g \) +H rSub { size 8{2} } O} {}

By measuring the mass of NaHCO 3 size 12{"NaHCO" rSub { size 8{3} } } {} and balancing the equation (above), the mass of NaCl expected to be produced can be calculated and then checked experimentally. Then, the actual amount of NaCl produced can be compared to the predicted amount.

This process includes molar ratios, molar masses, balancing and interpreting equations, and conversions between grams and moles and can be summarized as follows:

  • Check that the chemical equation is correctly balanced.
  • Using the molar mass of the given substance, convert the mass given in the problem to moles.
  • Construct a molar proportion (two molar ratios set equal to each other). Use it to convert to moles of the unknown.
  • Using the molar mass of the unknown substance, convert the moles just calculated to mass.

In the second part of this lab, since a great deal of chemistry is done with solutions, a solution will be prepared of allocated molarity. Molarity, or more correctly molar concentration, is defined to be the number of moles of solute divided by the number of liters of solution:

c M = n substance V solution size 12{c rSub { size 8{M} } = { {n rSub { size 8{ bold "substance"} } } over {V rSub { size 8{ ital "solution"} } } } } {}

with units of [mole/L]. However molar concentration depends on the temperature so a higher temperature would result in an increased volume with a consequential decrease in molar concentration. This can be a significant source of error, of the same order as the error in the volume measurements of a burette, when the temperature increases more than 5ºC.

Steps to preparing a solution of a certain concentration:

  • First need to know the formula for the solute, e.g. potassium chromate: K 2 CrO 4 size 12{K rSub { size 8{2} } "CrO" rSub { size 8{4} } } {} .
  • Need the molecular weight of the solute: by adding up the atomic weights of potassium, chromium and oxygen: 39.10, 52.00 and 16.00 in the correct ratios:
  • 2 × 39 . 1, 52 . 0 and 4 × 16 . 00 = 194 . 2g/mole . size 12{"2 " times " 39" "." "1, 52" "." "0 and 4 " times " 16" "." "00 = 194" "." "2g/mole" "." } {}
  • Then the volume of solution, usually deionised water: e.g. for one liter of solution use a 1000 mL volumetric flask. So a 1M solution would require 194.2g of solid K 2 CrO 4 size 12{K rSub { size 8{2} } "CrO" rSub { size 8{4} } } {} in 1 L, 0.1M 19.42g of solid K 2 CrO 4 size 12{K rSub { size 8{2} } "CrO" rSub { size 8{4} } } {} and so on.
  • Remember to ensure that all the solute is dissolved before finally filling to the mark on the volumetric flask. If there is any undissolved solute present in the solution, the water level will go down slightly below the mark, since the volume occupied by the solute differs from the actual volume it contributes to the solution once it is dissolved.

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Source:  OpenStax, Honors chemistry lab fall. OpenStax CNX. Nov 15, 2007 Download for free at http://cnx.org/content/col10456/1.16
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