Chemistry Lesson: Calculating Molar Mass, Percent Composition, and Empirical Formulas, Slides of Chemistry

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Announcements

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Molar Mass

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Percent Composition

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Empirical Formula

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Welcome to day 5!Chemical Formula

Announcements

•^

Seminar Today – 4:00 this room

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Information not covered in lecture are available elsewhere– Textbook– CAPA sets– CAPA tutors– Discussion sections– Office hours

Molar Mass

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Did this with thyroxine last time

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One more example – Heme–

C

32

H

26

FeN

O 4

4

–^

Just multiply and add

O

1mol

g (^999). 15

Heme mol 1

O

mol 4

N 1mol

g (^007). 14

Heme mol 1

N

mol 4

Fe 1mol

g (^845). 55

Heme mol 1

Fe mol 1

H mol 1

g (^0079). 1

Heme mol 1

H

mol 26 C mol 1

g (^011). 12

Heme mol 1

C

mol 32

× + × + × + × + ×

Heme 1mol

g (^996). 63

Heme 1mol

g (^028). 56

Heme 1mol

g (^845). 55

Heme mol 1

g (^2054). 26

Heme mol 1

g (^352). 384

Heme 586.4264gmol 1 =

Heme

586.43gmol 1

Note that C has 5 sig figs.After multiplying, 5 figs is hundredths place.Answer limited to hundredths place.

Percent Composition

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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

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Let’s stick with our heme example

Recall that the chemical formula for heme is C

32

H

26

FeN

O 4

4

O

1mol

g (^999). 15

Heme mol 1

O

mol 4

N 1mol

g (^007). 14

Heme mol 1

N

mol 4

Fe 1mol

g (^845). 55

Heme mol 1

Fe mol 1

H mol 1

g (^0079). 1

Heme mol 1

H

mol 26 C mol 1

g (^011). 12

Heme mol 1

C

mol 32

× + × + × + × + ×

5 4 3 2 1

Heme 1mol

g (^996). 63

Heme 1mol

g (^028). 56

Heme 1mol

g (^845). 55

Heme mol 1

g (^2054). 26

Heme mol 1

g (^352). 384

Heme

586.43gmol 1

What is the mass percent of nitrogen in heme? 0%0%0%^ 0%

10.913 %

9.3524 %

65.711 %

9.5541 %

Empirical Formula

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In reality experiments are performed to measure relativeamounts of components.

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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

Chemical Formula – glucose

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With additional information we can find true chemical formula

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For instance, maybe mass spectrum shows–

Total mass of glucose roughly 180 g/mol

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We know C:H:O must be 1:2:

g mol

g mol

g mol

g mol

x

x

x^

= × + × + ×

g mol

g mol

x^

x

• Empirical formula is

CH

O 2

• Chemical formula is six times larger:

C

H 6

12

O

6

Empirical Formula of Magnesium Oxide•^

Start with known mass of Mg in a crucible

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Add Oxygen (by burning)

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Determine mass of Mg

Ox^

y

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Subtract to get mass of O

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Example:– Mass of crucible

9.58 g

– Mass of crucible + Mg

10.47 g

– Mass of crucible + Mg + O

11.02 g

Chemical Formula – Mag Oxide

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All we know is Mg:O ratio is 1:

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Maybe molecules really come as Mg

O 6

6

or Mg

O^4

4

or …

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We need more information!

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Maybe we find mass spectrum data or something– Main peak is near 40 g/mol

g mol

g mol

g mol

x

x^

×

×

g mol

g mol

x^

x

Therefore chemical formula is

MgO

Today

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Go to Chem seminar

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Relax

By Monday

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Finish up CAPA set

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Read remainder of Chapter 3

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Also read the parts of Chapt 3 that you haven’t read yet

Remember: You are done with thehomework when you understand it!