Introduction to Chemistry: Conversions, The Mole, and Stoichiometry Study Guide

Chapter 5: Conversions

Scientific Notation

Scientific notation is a method for expressing very large or very small numbers in a compact form using powers of ten.

• Definition: A number is written as $a \times 10^n$, where $1 \leq a < 10$ and $n$ is an integer.

• Example: $0.00056 = 5.6 \times 10^{-4}$

Significant Figures

Significant figures are the digits in a measurement that are known with certainty plus one digit that is estimated.

• Rules: All nonzero digits are significant; zeros between nonzero digits are significant; leading zeros are not significant; trailing zeros in a decimal number are significant.

• Example: 0.00450 has three significant figures.

SI Units and Base Units

The International System of Units (SI) is the standard for scientific measurements.

• Base units: meter (m), kilogram (kg), second (s), mole (mol), ampere (A), kelvin (K), candela (cd)

• Example: Length is measured in meters (m).

Dimensional Analysis

Dimensional analysis is a technique for converting between units using conversion factors.

• Method: Multiply by conversion factors so that units cancel appropriately.

• Example: To convert 5.0 cm to meters: $5.0 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = 0.050 \text{ m}$

Metric Prefixes and Unit Conversions

Metric prefixes indicate powers of ten for SI units.

• Common prefixes: kilo (k, $10^3$), deci (d, $10^{-1}$), centi (c, $10^{-2}$), milli (m, $10^{-3}$), micro ($\mu$, $10^{-6}$), nano (n, $10^{-9}$)

• Example: 1 kilometer = $1 \times 10^3$ meters

Multistep Conversion Problems

Some problems require multiple conversion steps between units.

• Method: Set up a chain of conversion factors, canceling units at each step.

• Example: Convert 2.5 hours to seconds: $2.5 \text{ hr} \times \frac{60 \text{ min}}{1 \text{ hr}} \times \frac{60 \text{ s}}{1 \text{ min}} = 9000 \text{ s}$

Density Calculations

Density is the mass per unit volume of a substance.

• Formula: $\text{Density} = \frac{\text{mass}}{\text{volume}}$

• Units: g/mL, g/cm3

• Application: Water displacement can be used to measure the volume of irregular objects.

Chapter 6: The Mole

Moles and Avogadro's Number

The mole is the SI unit for amount of substance, defined as containing $6.022 \times 10^{23}$ entities (Avogadro's number).

• Definition: 1 mole = $6.022 \times 10^{23}$ particles (atoms, molecules, ions)

• Example: 1 mole of carbon atoms = $6.022 \times 10^{23}$ carbon atoms

Molar Mass of Elements and Compounds

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol).

• Calculation: For elements, use atomic mass from the periodic table; for compounds, sum the atomic masses of all atoms in the formula.

• Example: Molar mass of H2O = $2 \times 1.01$ g/mol + $16.00$ g/mol = $18.02$ g/mol

Conversions Involving Moles

Conversions between moles, mass, number of particles, and volume are fundamental in chemistry.

• Atoms, molecules, moles, and masses: Use Avogadro's number and molar mass as conversion factors.

• Example: To find the number of molecules in 3.0 moles of CO2: $3.0 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol} = 1.81 \times 10^{24} \text{ molecules}$

Compound and Constituent Element Conversions

Convert between moles of a compound and moles of its constituent elements using the chemical formula.

• Example: 1 mole of H2O contains 2 moles of H and 1 mole of O.

Grams and Moles of Constituent Elements

Calculate the mass of an element in a given mass of compound using molar ratios.

• Example: Find grams of H in 18.02 g of H2O: $18.02 \text{ g H}_2\text{O} \times \frac{2.02 \text{ g H}}{18.02 \text{ g H}_2\text{O}} = 2.02 \text{ g H}$

Mass Percent and Conversion Factor

Mass percent expresses the mass of an element as a percentage of the total mass of a compound.

• Formula: $\text{Mass percent} = \frac{\text{mass of element in 1 mol compound}}{\text{molar mass of compound}} \times 100\%$

• Application: Used to determine composition and for conversion calculations.

Empirical and Molecular Formulas

The empirical formula shows the simplest whole-number ratio of elements in a compound; the molecular formula shows the actual number of atoms of each element.

• Example: Empirical formula of glucose (C6H12O6) is CH2O; molecular formula is C6H12O6.

Chapter 7: Stoichiometry

Stoichiometric Conversions

Stoichiometry involves quantitative relationships between reactants and products in chemical reactions.

• Reactants and Products: Use balanced chemical equations to convert between moles of reactants and products.

• Example: $2H_2 + O_2 \rightarrow 2H_2O$; 2 moles of H2 produce 2 moles of H2O.

Mole and Mass Conversions

Convert between moles and masses of reactants and products using molar mass.

• Method: Mass $\rightarrow$ Moles $\rightarrow$ Moles (other substance) $\rightarrow$ Mass

• Example: How many grams of H2O are produced from 4.0 g H2? $4.0 \text{ g H}_2 \times \frac{1 \text{ mol H}_2}{2.02 \text{ g H}_2} \times \frac{2 \text{ mol H}_2O}{2 \text{ mol H}_2} \times \frac{18.02 \text{ g H}_2O}{1 \text{ mol H}_2O} = 35.7 \text{ g H}_2O$

Limiting Reactant and Yield Calculations

The limiting reactant is the reactant that is completely consumed first, limiting the amount of product formed.

• Theoretical yield: Maximum amount of product possible from given reactants.

• Actual yield: Amount of product actually obtained from the reaction.

• Percent yield: $\text{Percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$

• Excess reactant: Reactant that remains after the reaction is complete.

Stoichiometry Table

The following table summarizes key stoichiometric relationships:

Conversion	Factor Used	Example
Mass to Moles	Molar Mass	$\text{g} \rightarrow \text{mol}$
Moles to Moles	Balanced Equation Coefficients	$\text{mol A} \rightarrow \text{mol B}$
Moles to Mass	Molar Mass	$\text{mol} \rightarrow \text{g}$
Limiting Reactant	Compare moles of reactants	Find which reactant produces less product
Percent Yield	Actual/Theoretical Yield	$\frac{\text{actual}}{\text{theoretical}} \times 100\%$