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Take out #46 & stampsheet . Homework … Complete prelab. Do Now : If one stick of Juicy Fruit gum weighs 3.0g, what percent of the total mass of the gum is sugar?. 24.305. 35.453. Mg. Cl. 12. 17. magnesium. chlorine. Percentage Composition. (by mass...not atoms). 24.31 g 95.21 g.

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slide1

Take out #46 & stampsheet.

Homework …

Complete prelab

slide2

Do Now:

If one stick of Juicy Fruit gum weighs 3.0g, what percent of the total mass of the gum is sugar?

percentage composition

24.305

35.453

Mg

Cl

12

17

magnesium

chlorine

Percentage Composition

(by mass...not atoms)

24.31 g

95.21 g

% Mg = x 100

70.90 g

95.21 g

% Cl = x 100

25.53% Mg

Mg2+ Cl1-

74.47% Cl

MgCl2

It is not 33% Mg and 66% Cl

1 Mg @ 24.31 g= 24.31 g

2 Cl @ 35.45 g= 70.90 g

95.21 g

hydrate lab
Hydrate Lab

CuSO4 ∙ ?H2O

CuSO4

Hydrate Anhydrous

              • Water molecules are incorporated into the crystalline structure.
  • Hydrates, like zinc acetate dihydrate, Zn(C2H3O2)2 ∙ 2H2O are commonly found in skin care products (ex: moisturizer, shampoo & lip balm).
empirical and molecular formulas
Empirical and Molecular Formulas

A pure compound always consists of the same elements combined in the same proportions by weight.

Therefore, we can express molecular composition as PERCENT BY WEIGHT.

Ethanol, C2H6O

52.13% C

13.15% H

34.72% O

different types of formulas
Different Types of Formulas

C6H6

  • Molecular Formula – shows the real # of atoms in one molecule or formula unit
  • Empirical Formula – shows smallest whole number mole ratio

**Sometimes the empirical & molecular formula can be the same

  • Structural Formula- molecular formula info PLUS bonding electron and atomic arrangement

CH

the empirical formula march
The Empirical Formula March
    • Percent to grams
  • Grams to mole
  • Divide by smallest
  • Return to whole
empirical formula

50.04g C

5.59g H

44.37g O

/ 2.77 mol

/ 2.77 mol

/ 2.77 mol

Empirical Formula

Quantitative analysis shows that a compound contains 50.04% carbon, 5.59% hydrogen, and 44.37% oxygen.

Find the empirical formula of this compound.

= 1.5 C

= 4.17 mol C

50.04% C

5.59% H

44.37% O

  • x2
  • x2
  • x2

X

X

X

= 5.59 mol H

= 2 H

C3H4O2

= 2.77 mol O

= 1 O

Step 1) %  g

Step 2) g  mol

Step 3) mol

mol

Step 4) return

to whole

empirical formula1

66.75g Cu

10.84g P

22.41g O

/ 0.3500 mol

/ 0.3500 mol

/ 0.3500 mol

Empirical Formula

Quantitative analysis shows that a compound contains 66.75% copper, 10.84% phosphorus and 22.41% oxygen.

Find the empirical formula of this compound.

=3 Cu

66.75% Cu

10.84 % P

22.41 % O

= 1.050 mol Cu

X

X

X

Cu3PO4

= 1 P

= 0.3500 mol P

= 4 O

= 1.401 mol O

Step 1) %  g

Step 2) g  mol

Step 3) mol

mol

empirical formula2

32.38 g Na

22.65 g S

44.99 g O

/ 0.708 mol

/ 0.708 mol

/ 0.708 mol

Empirical Formula

Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen.

Find the empirical formula of this compound.

32.38% Na

22.65% S

44.99% O

X

X

X

= 2 Na

= 1.408 mol Na

Na2SO4

= 0.708 mol S

= 1 S

= 2.812 mol O

= 4 O

Step 1) %  g

Step 2) g  mol

Step 3) mol

mol