Theory - Unit 4 - The Mole Flashcards
Empirical Formula
Empirical Formula - A formula that gives the simplest whole-number ratio of atoms in a compound.
Dimensional Analysis
A method of mathmatical analysis which involves converting between units.
Percent Composition
The amount of a given element in a compound.
Scientific Notation
A way of expressing really large or really small numbers.
Negative Exponents in Scientific Notation
Negative Exponents in Scientific Notation mean you should move the decimal to the left to change the number to standard notation.
Turning Standard Notation into Scientific Notation
If the number is greater than 1, the exponent will be positive.
If the number is less than 1, the exponent will be negative.
The Mole
A unit of measure.
1 mole = the atomic mass of an element
and
1 mole = 6.02 E23 atoms, particles, molecules, units, etc.
Positive Exponents in Scientific Notation
Postive Exponents mean move the decimal to the right when changing a number to standard notation.
Significant Figures
All the digits in a measurement that are known with certainty plus a last digit that must be estimated.
Rules for Multiplying and Dividing with Significant Figures
The result has the same number of significant figures as the factor with the fewest significant figures.
Rules for Adding and Subtracting with Significant Figures
The result has the same number of decimal places as the number with the fewest decimal places.
How many atoms are in 2.7 moles of iron?
2.7 moles Fe x (6.02x10²³ atoms Fe/1 mole Fe) =
1.6 x 10²⁴atoms Fe
How many moles are in 3.67 x 10²⁴ molecules of SO₂?
- 67 x 10²⁴ molecules SO₂ x (1 mole SO₂/6.02x10²³ molecules SO₂)=
- 10 moles SO₂
9.08 g of Al₂0₃ is equal to how many moles?
9.08 g Al₂0₃ x (1 mole Al₂0₃/101.96 g Al₂0₃) = 0.0890 moles Al₂0₃
What mass of copper (II) sulfide is equal to 4.79 x 10²² atoms?
- 79 x 10²² atoms CuS x (1 mole CuS/6.02x10²³ atoms CuS) x (95.61 g CuS/1 mole CuS) =
- 61 g CuS
What is the percent composition of oxygen in CuSO₄?
mass of CuSO₄= 63.546 + 32.065 + 4(15.999) = 159.7 grams
Mass of oxygen = 4(15.999) = 64 grams
(64/159.7) x (100%) = 40.1. %O