take out lab calculations & stampsheet . homework … read 8.8 #’s 54-58, 75, 76

Take out lab calculations & stampsheet.

Homework …1.Read 8.82.#’s 54-58, 75, 76

Do Now:

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

% Cl = x 10070.90 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

25.53% Mg

74.47% Cl

(by mass...not atoms)

It is not 33% Mg and 66% Cl

% Mg = x 10024.31 g95.21 g

Hydrate Lab

HydrateAnhydrous

CuSO4 ∙ ?H2O CuSO4

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

Empirical and Molecular Empirical and Molecular FormulasFormulas

A pure compound always consists of the same elements combined in the same proportions by weight.Therefore, we can express

molecular composition as PERCENT PERCENT BY WEIGHTBY WEIGHT.Ethanol, C2H6O

52.13% C 13.15% H 34.72% O

Different Types of FormulasMolecular 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

The Empirical Formula The Empirical Formula MarchMarch

Percent to massPercent to mass Mass to moleMass to mole Divide by smallestDivide by smallest Return to wholeReturn to 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

C g 12.01

C mol 1

1 mol H

1.01 g H

1 mol O

16.00 g O

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

Step 4) return to whole

x2

x2

x2

X

X

X

Empirical FormulaQuantitative analysis shows that a compound contains 66.75% copper, 10.84% phosphorus and 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

1 mol P

30.97 g P

1 mol O

16 g O

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

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

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

X

X

X