Welcome to day 5!
• Announcements
• Molar Mass
• Percent Composition
• Empirical Formula
• Chemical Formula
Announcements
• Seminar Today – 4:00 this room
• Information not covered in lecture are available elsewhere
– Textbook
– CAPA sets
– CAPA tutors
– Discussion sections
– Office hours
Naming
• Be sure to examine rules in the text (pp. 73-76)
• Also see notes from lecture 4
• There will be exam questions requiring names throughout
the year
Molar Mass
•
•
Did this with thyroxine last time
One more example – Heme
– C32H26FeN4O4
– Just multiply and add
32mol C
12.011g
26mol H 1.0079g
1mol Fe
55.845g
4mol N
14.007g
4mol O
15.999g
×
+
×
+
×
+
×
+
×
1 mol Heme 1mol C 1 mol Heme 1mol H 1mol Heme 1mol Fe 1mol Heme 1mol N 1mol Heme 1mol O
=
384.352g
26.2054g
55.845g
56.028g
63.996g
+
+
+
+
1mol Heme 1mol Heme 1mol Heme 1mol Heme 1mol Heme
=
586.4264g
1mol Heme
=
586.43g
1mol Heme
Note that C has 5 sig figs.
After multiplying, 5 figs is hundredths place.
Answer limited to hundredths place.
Percent Composition
• Two types of percentage composition are logical
– Mole percent and Mass percent
• When text says “percent composition” they mean MASS
– For example, strange fruit salad ingredients
• 10 watermelons and 10 strawberries
– By moles, salad is 50 % watermelon
– By mass salad is 99 % watermelon
• Let’s stick with our heme example
C32H26FeN4O4 Composition
•
•
Mole percent
– We have 32+26+1+4+4 = 67 total moles of components
–
32
= 0.4776 ⇒ 47.8
67
Mole % Carbon (keep as many sig figs as you like)
–
1
= 0.0149 ⇒ 1.5
67
Mole % Iron
Mass percent
–
385.35
= 0.65711 ⇒ 65.711
586.43
Mass % Carbon (5 sig figs)
–
54.845
= 0.093524 ⇒ 9.3524
586.43
Mass % Iron (5 sig figs)
Recall that the chemical formula for heme is C32H26FeN4O4
32mol C
12.011g
26mol H 1.0079g
1mol Fe
55.845g
4mol N
14.007g
4mol O
15.999g
×
+
×
+
×
+
×
+
×
1 mol Heme 1mol C 1 mol Heme 1mol H 1mol Heme 1mol Fe 1mol Heme 1mol N 1mol Heme 1mol O
=
384.352g
26.2054g
55.845g
56.028g
63.996g
+
+
+
+
1mol Heme 1mol Heme 1mol Heme 1mol Heme 1mol Heme
=
586.43g
1mol Heme
What is the mass percent of nitrogen in heme?
0%
1. 10.913 %
0%
2. 9.3524 %
0%
3. 65.711 %
0%
4. 9.5541 %
1
2
3
4
5
Empirical Formula
• In reality experiments are performed to measure relative
amounts of components.
• For instance, analysis of glucose might yield:
– 40.000 % C
6.7136 % H
53.284 % O (by mass)
– Note that many forms of analysis are possible
– Text describes decomposition and combustion
• Read carefully!
– Let’s determine molar ratio of C:H:O in glucose
Empirical Formula – glucose
•
Assume some total mass of sample, say 100 g
40.000 g C ×
1 mol
= 3.3303 mol C
12.011 g
6.7136 g H ×
1 mol
= 6.6609 mol H
1.0079 g
53.284 g O ×
1 mol
= 3.3305 mol O
15.999 g
•
•
Since we simply chose initial mass, we don’t know absolute amounts
Only the relative mole amounts are relevant
•
Therefore, empirical formula is
CH2O
Chemical Formula – glucose
•
•
•
With additional information we can find true chemical formula
For instance, maybe mass spectrum shows
– Total mass of glucose roughly 180 g/mol
We know C:H:O must be 1:2:1
g
g
g
g
1x × 12.011 mol
+ 2 x × 1.0079 mol
+ 1x ×15.999 mol
= 180 mol
g
g
30.0258 x mol
= 180 mol
x = 5.9948
• Empirical formula is CH2O
• Chemical formula is six times larger:
C6H12O6
Empirical Formula of Magnesium Oxide
•
•
•
•
Start with known mass of Mg in a crucible
Add Oxygen (by burning)
Determine mass of MgxOy
Subtract to get mass of O
• Example:
– Mass of crucible
– Mass of crucible + Mg
– Mass of crucible + Mg + O
9.58 g
10.47 g
11.02 g
Empirical Formula2 – Mag Oxide
• Determine moles of Mg
– Mass of Mg = 10.47g – 9.58g = 0.89g
– Moles of Mg = 0.89g (1mol/24.305g) = 0.037 moles Mg
• Determine moles of O
– Mass of O = 11.02g – 10.47g = 0.55g
– Moles of O = 0.55g (1mol/15.999g) = 0.034 moles O
• Molar ratio?
– 0.37:0.34 = 1.1 ≈ 1
• Therefore empirical formula is MgO
Chemical Formula – Mag Oxide
• All we know is Mg:O ratio is 1:1
• Maybe molecules really come as Mg6O6 or Mg4O4 or …
• We need more information!
• Maybe we find mass spectrum data or something
– Main peak is near 40 g/mol
g
g
g
1x × 24.305 mol + 1x × 15.999 mol = 40 mol
g
g
40.304 x mol = 40 mol
x = 1.0076 ≈ 1
Therefore chemical formula is
MgO
Remember: You are done with the
homework when you understand it!
Today
• Go to Chem seminar
• Relax
By Monday
• Finish up CAPA set #3
• Read remainder of Chapter 3
• Also read the parts of Chapt 3 that you haven’t read yet