Counting by Weighing

Counting by Weighing

• Used for objects that are difficult or impossible to count individually.

• Involves weighing a group of items and using the average mass of a single item to determine the total number of items.

Example: Gummy Bears

• Scenario: Buying gummy bears in bulk.

• Problem: Counting each gummy bear individually is inefficient.

• Solution: Weigh the gummy bears and use the average mass to calculate the number of gummy bears.

Determining Average Mass

• Weigh a small number of gummy bears individually.

• Example: Weighing three gummy bears:

  • Gummy 1: 3 grams

  • Gummy 2: 4 grams

  • Gummy 3: 5 grams

• Average\ Mass = \frac{3 + 4 + 5}{3} = 4 \ grams

Conversion Factors

• Express the average mass as a mathematical equivalence:

  • 1 \ item = 4 \ grams

• Derive two conversion factors from this equivalence:

  • \frac{4 \ grams}{1 \ item}

  • \frac{1 \ item}{4 \ grams}

Calculating the Number of Items

• Weigh a larger sample of gummy bears.

• Example: A bag of gummy bears weighs 80 grams.

• Use the appropriate conversion factor to convert from grams to the number of items:

  • 80 \ grams * \frac{1 \ item}{4 \ grams} = 20 \ items

• Result: There are 20 gummy bears in the bag.

Application to Molecules

• Counting individual atoms or molecules is impossible.

• Use the concept of counting by weighing to determine the number of atoms or molecules in a sample.

Average Mass of a Carbon Atom

• The average mass of a carbon atom is 1.99366 \times 10^{-23} \ grams.

• The average mass accounts for the presence of isotopes.

Conversion factor for the average mass of Carbon atom

• Express the average mass as a mathematical equivalence:

  • 1 \ C \ atom = 1.99366 \times 10^{-23} \ grams

• Derive two conversion factors from this equivalence:

  • \frac{1.99366 \times 10^{-23} \ grams}{1 \ C \ atom}

  • \frac{1 \ C \ atom}{1.99366 \times 10^{-23} \ grams}

Significance of a large number of Atoms in a Visible Sample

• A sample of carbon that is visible to the naked eye contains a very large number of atoms because individual atoms are extremely small.

Atomic Mass Units (AMU)

• A new mass scale was introduced to express the mass of tiny particles without using scientific notation with huge negative exponents.

• Purpose: To express the mass of atoms in a more convenient way.

• Definition: 1 AMU = 1.66 \times 10^{-24} \ grams

• This definition is a math equivalence statement, and conversion factors can be derived. To convert between grams and AMUs.

• This is just gonna help us get between scales.

Conversions of AMU and Average mass example:

• Average mass of a carbon atom: 1.99366 \times 10^{-23} \ grams

• Convert the mass in grams to AMUs:

  • 1.99366 \times 10^{-23} \ grams * \frac{1 \ AMU}{1.66 \times 10^{-24} \ grams} = 12.01 \ AMU

Calculation using Scientific Notation

• Use the exponent key (EE or EXP) on a scientific calculator to enter numbers in scientific notation.

• Enter the number outside the exponent, press the exponent key, and then enter the exponent (including the negative sign, if applicable).

Atomic Mass on the Periodic Table

• Each element on the periodic table has an atomic mass listed.

• Example: Lithium (Li) has an atomic mass of 6.941.

• Interpretation: One lithium atom has a mass of 6.941 AMU.

• This number can be used to create conversion factors.

• This number is a math equivalent statement. We can take this statement and write our two conversions.

Calculations Using Atomic Mass

• Problem: How many atoms are in 56 AMU of carbon?

• Use the atomic mass of carbon (12.01 AMU) to convert from AMU to atoms:

  • 56 \ AMU * \frac{1 \ C \ atom}{12.01 \ AMU} = 4.66 \ C \ atoms

The Mole

• Just like AMU, it is an alternative that helps with small masses.

• The definition of a mole is that it's the number of carbon atoms in 12.01 grams of carbon.

• The explanation behind the definition has to do with the fact that atoms are very, very small and have very, very small masses.

• Avogadro's Number: 6.02 \times 10^{23}

Molar mass calculation to arrive at Avogadro's number

• \frac{12.01 \ grams}{1.9936610^{-23} \ grams} = 6.0210^{23}

Definition

• 1 \ mole = 6.02 \times 10^{23}

Molar Abbreviations

• The abbreviation of mole is mol.

Analogy with "Dozen"

• A mole is a packaging unit, just like a dozen (12).

• One mole of eggs = 6.02 \times 10^{23} eggs (an extremely large number).

• Moles are used to package tiny things like atoms and molecules.

Moles of atoms and molecules

• 1 \ mole = 6.02 \times 10^{23} atoms

• 1 \ mole = 6.02 \times 10^{23} molecules

Conversions

• \frac{1 \ mole \ of \ atoms}{6.02 \times 10^{23} \ atoms}

• \frac{1 \ mole \ of \ molecules}{6.02 \times 10^{23} \ molecules}

Molar Mass

• Definition: The mass of one mole of a substance.

• Units: grams/mole (g/mol)

• The atomic mass of an element (in AMU) is numerically equal to its molar mass (in grams/mole).

Conversions of Molar Mass

• \frac{grams}{mol}

Conversions between grams and mol:

• \frac{mol}{grams}

Example

• How many moles are in 2.45 grams of iron (Fe)?

• Use the molar mass of iron (55.85 g/mol) to convert from grams to moles:

  • 2.45 \ grams \ of \ Fe * \frac{1 \ mol \ of \ Fe}{55.85 \ grams \ of \ Fe} = 0.0439 \ mol \ of \ Fe
    *How many atoms are in 2.45 grams of iron (Fe)?

• Use the molar mass and then Avogadro's number:

  • 0.0439 \ mol \ of \ Fe * \frac{6.0210^{-23} \ atoms \ of \ Fe}{1 \ mol \ of \ Fe} = 2.6410^{-22} \ atoms \ of \ Fe

Multiple Step Conversions

• How many atoms are in 66.9 grams of copper (Cu)?

• Grams -> Moles -> Atoms

• Use molar mass to conver grams to moles

• Use Avogadro's number to convert moles to atoms.

Conversions including metric prefixes

• How many milligrams are there in 6.19 \times 10^{20} atoms of magnesium (Mg)?

• Atoms -> Moles -> Grams -> Milligrams

• The above is a three step conversion. You can convert atoms to moles with Avogadro's number. Moles to grams with molar mass, then grams to milligrams using metric prefixes.

Calculating Molar Mass of Molecules

• Find the mass of each element, multiply with the number of atoms.

• Add all the masses for the molar mass of the full molecule.

Molar mass calculation

• Using B2H6 as an example we see that there are 2 moles of baron at 10.81 grams per mole. Then there are six moles of hydrogen with 1.01 grams per mole. Add all of that together to get 27.68 grams. Molar mass is 27.68 grams per mole. Units g/mol. Even though we have an equivalent conversion here on the left, no one is gonna say the molar mass of p two h six is moles per 27.68 grams

Structural Formulas

• There can be more complex formulas to give a better understanding of the bonds occurring.

Aluminum Nitrate calculation

• When computing the calculation, the Aluminum molar mass is 26.98. There are three moles of nitrogen so the molar mass is 42.03. THen there are 9 moles of oxygen so the molar mass is 144. Adding together we get that the total molar mass of aluminum nitrate is 213.01 grams per mole.

Molar Mass with Dimensional Analysis

• Starting with the question of how many molecules are in 2.87 grams of methane the first thing to calculate is the molar mass of the molecule.

• Methane mass is 16.05 grams per mole.

• Following is the setup for a 2 step multiple chain dimensional analysis calculation.

Multiple Chain Dimensional Analysis Calculation

• 2.87 \ grams \ of \ CH4 * \frac{1 \ mol \ of \ CH4}{16.05 \ grams \ of \ CH4} * \frac{6.0210^{-23} \ molecules \ of \ CH4}{1 \ mol \ of \ CH4} = 1.0810^{-23} \ molecules \ of \ CH4

Formulas for % composition

• The mass precent formula will be on your exam, however what isn't on there is that everything is must be in mass. Remember each number you produce, should be in mass units. So be cautious of units. If you aren't in mass it is easy to mess up. The ratio must be, element mass contributed divided by total molar mass compounded.

Percent Composition Calculations

• A quick and easy trick is to make sure that mass percent sums approximately 100. In our notes here a quick way to check is to complete the calculation manually and then verify. Rounding can sometimes cause incorrect or misleading information leading to a large delta in % offset relative to 100%.

• During the calculations on the slides you should show the appropriate units. Molar mass should be given as mass per mass units. If it is missing it is often easy to miss. I need to see to get so what I do is usually just read left to right. You understand it doesn't matter what order we add them together.

• Remember always show your units and always practice the problems. If they are helpful make sure to use them, as they will be useful on Wednesday. When calculating there is always a quick way to check on tests etc, so if you practice you'll find a pattern.