Download presentation

Presentation is loading. Please wait.

Published byTomas Pay Modified over 3 years ago

1
What Could It Be? Finding Empirical and Molecular Formulas

2
**It’s kind of like a fraction reduced to its lowest terms.**

Empirical Formulas The empirical formula is the simplest whole number ratio of the atoms of each element in a compound. Note: it is not necessarily the true formula of the compound. Example: The molecular formula for glucose is C6H12O6, but its empirical formula is CH2O It’s kind of like a fraction reduced to its lowest terms.

3
Molecular Formulas The molecular formula gives the actual numbers of each element, and thereby represents the true formula of the compound.

4
**Calculating the Empirical Formula**

Example 1: A compound is found to contain the following… 2.199 g copper 0.277 g oxygen Calculate its empirical formula.

5
**Calculating the Empirical Formula**

Step 1: Convert the masses to moles. Copper: Oxygen:

6
**Calculating the Empirical Formula**

Step 2: Divide all the moles by the smallest value This gives the “mole ratio”

7
**Calculating the Empirical Formula**

Step 3: Round off these numbers, they become the subscripts for the elements. Cu2O

8
**Example 2: A material is found to be composed of 38. 7% carbon, 51**

Example 2: A material is found to be composed of 38.7% carbon, 51.6% oxygen, and 9.7% hydrogen. By other means, it is known that the molecular weight is 62.0 g/mol Calculate the empirical and molecular formulas. If you assume a sample weight of 100 grams, then the percents are really grams.

9
**Now, divide all the moles by the smallest one, 3.22**

Example 2: A material is found to be composed of 38.7% carbon, 51.6% oxygen, and 9.7% hydrogen. By other means, it is known that the molecular weight is 62.0 g/mol Calculate the empirical and molecular formulas. Carbon: Now, divide all the moles by the smallest one, 3.22 Oxygen: Hydrogen:

10
**Example 2: A material is found to be composed of 38. 7% carbon, 51**

Example 2: A material is found to be composed of 38.7% carbon, 51.6% oxygen, and 9.7% hydrogen. By other means, it is known that the molecular weight is 62.0 g/mol Calculate the empirical and molecular formulas. So, the empirical formula must be CH3O The molecular weight of the empirical formula is…. C x = g/mol H x 3 = g/mol O x 1 = g/mol 31.04 g CH3O/mol CH3O

11
**Remember, the empirical formula is not necessarily the molecular formula!**

MW of the empirical formula = 31.04 MW of the molecular formula = 62.0

12
2 x ( ) CH3O = C2H6O2 Empirical Formula 31.04 g/mol Molecular Formula 62.0 g/mol

13
**Remember, the molecular formula represents the actual formula.**

14
**What if the mole ratios don’t come out even?**

15
**Example 3: A compound is analyzed and found to contain 2**

Example 3: A compound is analyzed and found to contain 2.42 g aluminum and 2.15 g oxygen. Calculate its empirical formula. Aluminum: X 2 = 2 Oxygen: X 2 = 3 Al2O3

Similar presentations

OK

10-3: Empirical and Molecular Formulas. Percentage Composition The mass of each element in a compound, compared to the mass of the entire compound (multiplied.

10-3: Empirical and Molecular Formulas. Percentage Composition The mass of each element in a compound, compared to the mass of the entire compound (multiplied.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on life insurance in india Ppt on non conventional sources of energy Ppt on online mobile shopping Ppt on job rotation example Download ppt on pulse code modulation video Ppt on sources of energy for class 8th Download ppt on cell organelles and their functions Ppt on conservation of momentum equation Ppt on power system harmonics study Easy ppt on bluetooth