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

Hydrate Lab – for tomorrow

Pre-lab Title Purpose Materials Procedure Data table

Hydrate Anhydrous

Hydrates

Methane hydrate

Hydrates, like zinc acetate dihydrate, Zn(C2H3O2)2 * 2H2O are commonly found in skin care products such as moisturizer, shampoo and lip balm.

What is the % H2O in nickel chloride dihydrate, NiCl2 * 2H2O?

Element # g/mol (molar mass)

TOTAL

Ni 1 58.69 g/mol 58.69g

Cl 2 35.45 g/mol 70.90g

H2O 2 18.02 g/mol 36.04g

MOLAR MASS=

165.63g/mol NiCl2 *

2H2O

€

36.04 g H2O

165.63 g NiCl2 * 2H2O

⎛

⎝ ⎜

⎞

⎠ ⎟%H2O = = 21.76% H2O

% Mg = x 10024.31 g95.21 g

Percentage CompositionPercentage Composition

Mg

magnesium

24.305

12Cl

chlorine

35.453

17

Mg2+ Cl1-

MgCl2

1 Mg @ 24.31 g= 24.31 g2 Cl @ 35.45 g= 70.90 g 95.21 g

25.52% Mg

74.48% Cl

(by mass...not atoms)

It is not 33% Mg and 66% Cl

% = x 100part

whole

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 WEIGHTPERCENT BY WEIGHT.

Ethanol, C2H6O 52.13% C 13.15% H 34.72% O

Different Types of Formulas 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

C6H6

CH

Calculating Empirical formula

Percent to mass (assume 100g) Mass to mole (molar mass) Divide by the smallest (ratio)

Multiply til’ whole*

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

Find the empirical formula of this compound.

= 4.17 mol C

= 5.59 mol H

= 2.77 mol O

/ 2.77 mol

/ 2.77 mol

/ 2.77 mol

= 1.5 C

= 2 H

= 1 O

C3H4O2

50.04% C

5.59% H

44.37% O

50.04g C

5.59g H

44.37g O

€

1 mol C

12.01 g C

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⎝ ⎜

⎞

⎠ ⎟

€

1 mol H

1.01 g H

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⎝ ⎜

⎞

⎠ ⎟

€

1 mol O

16.00 g O

⎛

⎝ ⎜

⎞

⎠ ⎟

Step 1) % g Step 2) g mol Step 3) mol mol

Step 4) multiply til whole

*2

*2

*2

X

X

X

Empirical FormulaQuantitative analysis shows that a compound contains 66.75% copper, 10.84% phosphorusand 22.41% oxygen.

Find the empirical formula of this compound.

= 1.050 mol Cu

= 0.3500 mol P

= 1.401 mol O

/ 0.3500 mol

/ 0.3500 mol

/ 0.3500 mol

=3 Cu

= 1 P

= 4 O

Cu3PO4

66.75% Cu

10.84 % P

22.41 % O

66.75g Cu

10.84g P

22.41g O

€

1 mol Cu

63.55 g Cu

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⎝ ⎜

⎞

⎠ ⎟

€

1 mol P

30.97 g P

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⎝ ⎜

⎞

⎠ ⎟

€

1 mol O

16 g O

⎛

⎝ ⎜

⎞

⎠ ⎟

copper (I) phosphate

Step 1) % g Step 2) g mol Step 3) mol mol

Cu3PO4

X

X

X

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

Find the empirical formula of this compound.

= 1.408 mol Na

= 0.708 mol S

= 2.812 mol O

/ 0.708 mol

/ 0.708 mol

/ 0.708 mol

= 2 Na

= 1 S

= 4 O

Na2SO4

32.38% Na

22.65% S

44.99% O

32.38 g Na

22.65 g S

44.99 g O

Na g 23

Na mol 1

S g 32

S mol 1

O g 16

O mol 1

sodium sulfate

Step 1) % g Step 2) g mol Step 3) mol mol

Na2SO4

X

X

X