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Chemistry 121 Formulas and Constants

Chapter 1

Fahrenheit (K): \dfrac{9}{5}\left( °C+32\right)

Celsius (°C): \dfrac{5}{9}\left( °F-32\right)

Kelvin (K): °C+273

Kelvin → Celsius: K-273

Density: \dfrac{m}{V}

Units of Measurement

Physical Quantity

Name of Units

Abbreviation

Mass

gram

g

Length

meter

m

Time

seconds

s

Temperature

Celsius or Kelvins

°C or K

Amount of a substance

mole

mol

Volume

cubic centimeter or liter

cm³ or l


Chapter 2

1 amu = 1.66054\cdot 10^{-24}\ g

Atomic weight = \Sigma [(mass)(% fractional abundance)] for ALL isotopes

  • Percentage to decimal, multiply by 100

  • Decimal to percentage, divide by 100

  • Example of set up: amu = (mass)(% fractional abundance) + (mass)(% fractional abundance) + (mass)(% fractional abundance)

    • 3 total isotopes in this example


Chapter 3

Percent composition (% element) = [(# of atoms)(molecular weight)/ molar mass of the compound] x 100

Avogadro’s number = 6.02\times 10^{23} atoms or molecules

Percent yield = (actual yield/theoretical yield) x 100%

I will slowly add other chapters

S

Chemistry 121 Formulas and Constants

Chapter 1

Fahrenheit (K): \dfrac{9}{5}\left( °C+32\right)

Celsius (°C): \dfrac{5}{9}\left( °F-32\right)

Kelvin (K): °C+273

Kelvin → Celsius: K-273

Density: \dfrac{m}{V}

Units of Measurement

Physical Quantity

Name of Units

Abbreviation

Mass

gram

g

Length

meter

m

Time

seconds

s

Temperature

Celsius or Kelvins

°C or K

Amount of a substance

mole

mol

Volume

cubic centimeter or liter

cm³ or l


Chapter 2

1 amu = 1.66054\cdot 10^{-24}\ g

Atomic weight = \Sigma [(mass)(% fractional abundance)] for ALL isotopes

  • Percentage to decimal, multiply by 100

  • Decimal to percentage, divide by 100

  • Example of set up: amu = (mass)(% fractional abundance) + (mass)(% fractional abundance) + (mass)(% fractional abundance)

    • 3 total isotopes in this example


Chapter 3

Percent composition (% element) = [(# of atoms)(molecular weight)/ molar mass of the compound] x 100

Avogadro’s number = 6.02\times 10^{23} atoms or molecules

Percent yield = (actual yield/theoretical yield) x 100%

I will slowly add other chapters

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