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Do Now : 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. Water molecules are incorporated into the crystalline structure.

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slide1

Do Now:

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
Hydrate Lab – for tomorrow
  • Pre-lab
    • Title
    • Purpose
    • Materials
    • Procedure
    • Data table

Hydrate Anhydrous

hydrates
Hydrates
  • Water molecules are incorporated into the crystalline structure.

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.
percentage composition

part

whole

24.31 g

95.21 g

% = x 100

% Mg = x 100

24.305

35.453

Mg

Cl

12

17

magnesium

chlorine

Percentage Composition

(by mass...not atoms)

25.52% Mg

Mg2+ Cl1-

74.48% 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

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

calculating empirical formula
Calculating Empirical formula
  • Percent to mass (assume 100g)
  • Mass to mole (molar mass)
  • Divide by the smallest (ratio)
  • Multiply til’ whole*
empirical formula

/ 2.77 mol

50.04g C

5.59g H

44.37g O

/ 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

X

X

X

50.04% C

5.59% H

44.37% O

= 4.17 mol C

  • *2
  • *2
  • *2

C3H4O2

= 5.59 mol H

= 2 H

= 2.77 mol O

= 1 O

Step 1) %  g

Step 2) g  mol

Step 3) mol

mol

Step 4) multiply

til whole

empirical formula1

/ 0.3500 mol

66.75g Cu

10.84g P

22.41g O

/ 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.

copper (I) phosphate

=3 Cu

X

X

X

66.75% Cu

10.84 % P

22.41 % O

= 1.050 mol Cu

Cu3PO4

Cu3PO4

= 0.3500 mol P

= 1 P

= 1.401 mol O

= 4 O

Step 1) %  g

Step 2) g  mol

Step 3) mol

mol

empirical formula2

/ 0.708 mol

32.38 g Na

22.65 g S

44.99 g O

/ 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.

sodium sulfate

= 2 Na

X

X

X

32.38% Na

22.65% S

44.99% O

= 1.408 mol Na

Na2SO4

Na2SO4

= 0.708 mol S

= 1 S

= 2.812 mol O

= 4 O

Step 1) %  g

Step 2) g  mol

Step 3) mol

mol