Mole and Mass Relationships
Nick DiPreta
Jonny Masci
Chemistry
Due: 3/1/10
Pre-Lab Discussion:
In a balanced equation, all reactants and products must be represented by symbols or formulas.
The total number of atoms on each side of the equation must be equal on each side of the
equation, in accordance with the conservation of mass.
Thus, to calculate the formula mass of a reactant or product allows a researcher to convert
from grams of a certain substance taking part in a reaction to moles. The mole relationship or
ratio shown by the coefficients of the balanced equation allows the researcher to calculate how
many mole of all the other substances will take part in the reaction.
In the experiment that preceded, we investigated the relationships in the reaction
NaHCO3 (s) + HCl (aq)  NaCl(aq)+ CO2 (g) + H20 (g)
A known mass of the sodium hydrogen carbonate will react with excess hydrochloric acid. By
knowing the mass of the NaHCO3 that reacts, one can determine from the balanced equation
not only the mass of NaCl that should be produced, but one can also compare this value with
the experimental value that was produced.
Purpose
The experiment aided in the understanding of the mole-mass relationships that exist in a
chemical reaction and in the interpretation of a balanced chemical equation. The purpose of
this experiment was to compare the experimental mass of a product of a chemical reaction
with the predicted mass for that product.
Equipment/Materials: balance, burner, evaporating dish, watch glass, microspatula, ring stand,
iron ring, wire gauze, safety goggles, lab apron/coat, 6M hydrochloric acid, sodium hydrogen
carbonate
Safety
- be careful around the hydrochloric acid
-
-Wear safety goggles/lab apron
do not lean over the apparatus when heating it
note all caution alert symbols under “procedure”
Procedure:
1. Flame dry a clean evaporating dish by heating it in the hot part of a burner flame for
about 5 minutes. Allow the dish to cool.
2. Find the combined mass of the evaporating dish plus a watch glass. This mass is “a” in
your list of data.
3. Leaving the watch glass and evaporating dish on the balance, move the riders to
measure an additional 2.50 grams. Using a microspatula, add sodium hydrogen
carbonate to the evaporating dish until the scale balances. Record this mass as “b” in
your list of data.
4. Set up the ring stand, ring, and wire gauze as shown in figure 16-1. Place the watch glass on top
of the evaporating dish and place the dish on the wire gauze
5. Obtain about 5 mL of 6M hydrochloric acid (HCl) in a clean, dry test tube. Caution: Handle this
acid carefully. Slowly add the HCl to the NaHCO3 in the evaporating dish back and forth a couple
of times to make sure that the acid has contacted all the NaHCO3. After making sure that all
bubbling has stopped, remove the watch glass and place it curved side up on the lab bench.
6. Holding the burner in your hand, gently heat the evaporating dish. Use a low flame and move
the burner back and forth so as to avoid spattering. When almost all the liquid is gone, remove
the burner and replace the watch glass on the evaporating dish, leaving a small opening for
vapor to escape. Heat gently again until no liquid remains. Allow the dish to cool.
7. Find the combined mass of the watch glass, evaporating dish, and contents (NaCl). Record this
mass “C”, in your list of data.
Observations and Data
a. evaporating dish + watch glass
94.59 grams
b. evaporating dish + watch glass+ NaHCO3 97.59 grams
c. evaporating dish + watch glass + NaCl
96.65
Calculations
1. Find the mass of the NaHCO3 reactant, b-a.
97.59 g(evaporating dish + watch glass+ NaHCO3) – 94.59 g( evaporating dish + watch
glass) = 3 grams NaHCO3
2. Find the mass of the NaCl product, c-a.
96.65g(evaporating dish + watch glass + NaCl )- 94.59g (evaporating dish + watch glass)=
2.6 grams NaCl
Conclusions/Questions
1. According to the balanced equation for the reaction used in this experiment, what is the
ratio of moles of NaHCO3 reacted to moles of NaCl produced?
According to the balanced equation, the ratio of moles NaHCO3 reacted to moles of NaCl
produced is 1:1.
2. How many moles of NaHCO3 is reacted in this experiment? How many moles of NaCl is
produced? What is the ratio of moles NaHCO3 reacted to moles NaCl produced?
Formula mass NaHCO3 = 23 (Na) + 1 (H) + 12 (C) + 48 (O3)= 84g NaHCO3/mol
3.0 grams NaHCO3/ 84g NaHCO3 = 0.036 mol
Formula mass NaCl = 23 (Na)+ 35 (Cl)= 58 g/mol NaCl
2.6 grams NaCl/58 g/mol NaCl= 0.045 mol
0.036 mol NaHCO3/ 0.36 NaHCO3=1
0.045 mol NaCl /0.036 NaHCO3 = 1.25
The real ratio of NaHCO3 reacted to moles NaCl produced is 1:1, however the experimental
value obtained was 1:1.25
3. Using the balanced equation, calculate the mass of NaCl you would expect to get when 3.00
grams of NaHCO3 is reacted with HCl. How does this value compare with the
mass attained experimentally?
3.0 grams NaHCO3/ 84g NaHCO3 = 0.036 mol NaHCO3
0.036 X 58 g/mol NaCl= 2.08 grams NaCl
When I calculated the mass of NaCl I should have gotten when the 3.0 grams NaHCO3 is reacted
with HCl I got a value of 2.08 grams NaCl, which is over half a gram less than the
value I obtained.
4. If the masses of all but one of the substances that take part in a chemical reaction are
known, explain why it is possible to determine the unknown mass by
subtraction.
The law of conservation of mass states that matter in a chemical reaction can be neither
created, nor destroyed, thus if all but one of the substances that take part in a
chemical reaction are known, you can subtract the last mass to find out where
the rest of the mass has gone as there can be no more or less mass.
5. In the chemical reaction CaCO3  CaO+ CO2, if 40.0 g of CaCO3. Is decomposed:
a. how many grams of CaO is produced? 22.4 grams
b. how many grams of CO2 is produced? 17.6 grams
a. CaCO3=100 g/mol 40g/100g/mol= 0.4mol CaO/CaCO3/ CO2
0.4 mol CaO X 56g/mol CaO= 22.4 grams CaO
0.4 mol CO2 X 44g/mol CO2= 17.6 grams CO2
6. In the reaction N2+3H2 2NH3, if 20.0 grams of hydrogen reacts:
a. how many grams of ammonia is produced?113.22 grams
b. how many grams of nitrogen reacts?93.24 grams nitrogen
20.0 grams H2/ 2g/mol= 10 mol H2
10 mol H2 times (2mol ammonia/3mol H2)= 6.66 mol ammonia
6.66mol X 17 (molar mass ammonia)= 113.22 grams ammonia
10 mol H2 times (1mol nitrogen/ 3mol H2)= 3.33 mol nitrogen
3.33mol nitrogen X 28 (molar mass Nitrogen)=93.24 grams nitrogen
Conclusion:
This experiment aided in the understanding of the mole-mass relationships that exist in a
chemical reaction and in the interpretation of a balanced chemical equation. We can compare
the experimental mass of a product of a chemical reaction with the predicted mass for that
product. It shows that experimental values are not always precise.