Chapter 3 Atoms: the Building Blocks of Matter

1) Protons (p+) positively charged particle The parts that make up an atom are called subatomic particles. 1) Protons (p+) positively charged particle 2) Neutron (no) neutral particle (uncharged) 3) Electrons (e-) negatively charged particle Neutrons and Protons are located in the nucleus of an atom and are called hadrons. Electrons orbit around the nucleus.

Q- How are atoms of different elements distinguished from one another? In other words, how do we distinguish a helium atom from a carbon atom? A- Their number of protons, indicated by the atomic number Let’s look at helium, He. It has an atomic number of 2, which means that is has 2 protons in it’s nucleus.

Here are the basics; you need to know these. Atomic Structure Here are the basics; you need to know these. 1 H 1.0076 Hydrogen Atomic Number (Z): the number of protons (p+) Atomic Mass: the number of protons (p+) + the number of neutrons (n0) ▪ measured in atomic mass units (amu) which is one twelfth the mass of a carbon-12 atom. ▪ the mass of electrons (1/1860 p+) is negligible. Number of Neutrons: the atomic mass - the atomic number Atomic Number Atomic Symbol Atomic Mass

Lets practice! Find the missing information? Element Atomic # Atomic Mass Protons Electrons Neutrons Ar 18 39.948 amu He 2 O 15.999 amu 8

The Famous Gold Foil Experiment This showed us that the atom is made of mostly empty space.

Isotopes Atoms of the same element with different number of neutrons Because they have the same number of protons, all isotopes of an element have the same chemical properties.

Mass Numbers of Hydrogen Isotopes What would the masses be?

Calculating Average Atomic Mass Look at the mass number of any given element, and you will notice that it is a rather unruly decimal.

The Mole: A Measurement of Matter At the end of this section, you should be able to: Describe how Avogadro’s number is related to a mole of any substance Calculate the mass of a mole of any substance

The Mole (aka Avagadro’s Number): 6.02 x 1023

The Mole and Avogadro’s Number SI unit that measures the amount of substance 1 mole = 6.02 x 1023 representative particles Representative particles are usually atoms, molecules, or formula units (ions)

So, 1 mole of any substance is a set number of But Why the Mole? Just as 12 = 1 dozen, or 63,360 inches = 1 mile, the mole allows us to count microscopic items (atoms, ion, molecules) on a macroscopic scale. So, 1 mole of any substance is a set number of Items, namely: 6.02 x 1023. Chemistry = awesome

Representative Particle Representative Particles in 1.00 mol Examples: Substance Representative Particle Chemical Formula Representative Particles in 1.00 mol Atomic nitrogen Atom N 6.02 x 1023 Water Molecule H2O Calcium ion Ion Ca2+

Representative Particles in 1.00 mol Solve Substance Representative Particle Formula Unit Representative Particles in 1.00 mol Nitrogen gas N2 Molecule Calcium Fluoride CaF2 Sucrose C12H22O11 Carbon C Atom

All have 6.02 x 1023 representative particles in 1.00 mol Answers Nitrogen gas-molecule-N2 Calcium fluoride-molecule-CaF2 Sucrose-molecule-C12H22O11 Carbon-atom-C All have 6.02 x 1023 representative particles in 1.00 mol

How many atoms are in a mole? Determined from the chemical formula List the elements and count the atoms Solve for CO2 C - 1 carbon atom O - 2 oxygen atoms Add: 1 + 2 = 3 Answer: 3 times Avogadro’s number of atoms

Solve: How many atoms are in a mole of 1. Carbon monoxide – CO 2. Glucose – C6H12O6 3. Propane – C3H8 4. Water – H2O

How many moles of magnesium is 1.25 x 1023 atoms of magnesium? Divide the number of atoms or molecules given in the example by 6.02 x 1023 Divide (1.25 x 1023) by (6.02 x 1023) Express in scientific notation Answer = 2.08 x 10-1 mol Mg

Moles and Atoms Conversions Converting moles to representative particles: Converting representative particles to moles:

Objectives Use the molar mass to convert between mass and moles of a substance Use the mole to convert among measurements of mass, volume, and number of particles

Molar mass Mass (in grams) of one mole of a substance Broad term (can be substituted) for gram atomic mass, gram formula mass, and gram molecular mass Can be unclear: What is the molar mass of oxygen? O or O2 ? - element O or molecular compound O2 ?

Molar Mass Gram atomic mass (gam) – atomic mass of an element taken from the periodic table Gram molecular mass (gmm) – mass of one mole of a molecular compound Gram formula mass (gfm) – mass of one mole of an ionic compound Can use molar mass instead of gam, gmm, or gfm

Calculating the Molar Mass of Compounds (Molecular and Ionic) 1. List the elements 2. Count the atoms 3. Multiply the number of atoms of the element by the atomic mass of the element (atomic mass is on the periodic table) 4. Add the masses of each element 5. Express to tenths place

What is the molar mass (gfm) of ammonium carbonate (NH4)2CO3? N 2 x 14.0 g = 28.0 g H 8 x 1.0 g = 8.0 g C 1 x 12.0 g = 12.0 g O 3 x 16.0 g = 48.0 g Add ________ Answer 96.0 g

Practice Problems 1. How many grams are in 9.45 mol of dinitrogen trioxide (N2O3) ? a. Calculate the grams in one mole b. Multiply the grams by the number of moles 2. Find the number of moles in 92.2 g of iron(III) oxide (Fe2O3). b. Divide the given grams by the grams in one mole

Answers 1. 718 g N2O3 (one mole is 76.0g) 2. 0.578 mol Fe2O3 (one mole is 159.6 g)

Volume of a Mole of Gas Varies with a change in temperature or a change in pressure At STP, 1 mole of any gas occupies a volume of 22.4 L Standard temperature is 0°C Standard pressure is 101.3 kPa (kilopascals), or 1 atmosphere (atm) 22.4 L is known as the molar volume

22. 4 L of any gas at STP contains 6 22.4 L of any gas at STP contains 6.02 x 1023 representative particles of that gas One mole of a gaseous element and one mole of a gaseous compound both occupy a volume of 22.4 L at STP (Masses may differ) Study Figure 7.13 on page 186 Molar mass (g/mol) = Density (g/L) x Molar Volume (L/mol)

Objectives Define the terms Calculate the percent composition of a substance from its chemical formula or experimental data Derive the empirical formula and the molecular formula of a compound from experimental data

Terms to Know Percent composition – relative amounts of each element in a compound Empirical formula – lowest whole- number ratio of the atoms of an element in a compound

An 8. 20 g piece of magnesium combines completely with 5 An 8.20 g piece of magnesium combines completely with 5.40 g of oxygen to form a compound. What is the percent composition of this compound? 1. Calculate the total mass Divide each given by the total mass and then multiply by 100% Check your answer: The percentages should total 100%

Answer The total mass is 8.20 g + 5.40 g = 13.60 g Divide 8.2 g by 13.6 g and then multiply by 100% = 60.29412 = 60.3% Divide 5.4 g by 13.6 g and then multiply by 100% = 39.70588 = 39.7% Check your answer: 60.3% + 39.7% = 100%

1) Find the percent composition of Aluminum Oxide (Al3O2) 2) How much of a 5-g piece of Iron Bromide (FeBr3) is iron?

Calculate the percent composition of propane (C3H8) 1. List the elements 2. Count the atoms 3. Multiply the number of atoms of the element by the atomic mass of the element (atomic mass is on the periodic table) 4. Express each element as a percentage of the total molar mass 5. Check your answer

Answer Total molar mass = 44.0 g/mol 36.0 g C = 81.8% 8.0 g H = 18.2%

Calculate the mass of carbon in 52.0 g of propane (C3H8) Calculate the percent composition using the formula (See previous problem) 2. Determine 81.8% of 82.0 g Move decimal two places to the left (.818 x 82 g) 3. Answer = 67.1 g

Calculating Empirical Formulas Microscopic – atoms Macroscopic – moles of atoms Lowest whole-number ratio may not be the same as the compound formula Example: The empirical formula of hydrogen peroxide (H2O2) is HO

Empirical Formulas The first step is to find the mole-to-mole ratio of the elements in the compound If the numbers are both whole numbers, these will be the subscripts of the elements in the formula If the whole numbers are identical, substitute the number 1 Example: C2H2 and C8H8 have an empirical formula of CH If either or both numbers are not whole numbers, numbers in the ratio must be multiplied by the same number to yield whole number subscripts

What is the empirical formula of a compound that is 25 What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen? 1. Assume 100 g of the compound, so that there are 25.9 g N and 74.1 g O 2. Convert to mole-to-mole ratio: Divide each by mass of one mole 25.9 g divided by 14.0 g = 1.85 mol N 74.1 g divided by 16.0 g = 4.63 mol O 3. Divide both molar quantities by the smaller number of moles

4. 1.85/1.85 = 1 mol N 4.63/1.85 = 2.5 mol O 5. Multiply by a number that converts each to a whole number (In this case, the number is 2 because 2 x 2.5 = 5, which is the smallest whole number ) 2 x 1 mol N = 2 2 x 2.5 mol O = 5 Answer: The empirical formula is N2O5

Determine the Empirical Formulas 1. H2O2 2. CO2 3. N2H4 4. C6H12O6 5. What is the empirical formula of a compound that is 3.7% H, 44.4% C, and 51.9% N?

Answers Compound Empirical Formula 1. H2O2 HO 2. CO2 CO2 3. N2H4 NH2 4. C6H12O6 CH2O 5. HCN

Calculating Molecular Formulas The molar mass of a compound is a simple whole-number multiple of the molar mass of the empirical formula The molecular formula may or may not be the same as the empirical formula

Calculate the molecular formula of the compound whose molar mass is 60 Calculate the molecular formula of the compound whose molar mass is 60.0 g and empirical formula is CH4N. 1. Using the empirical formula, calculate the empirical formula mass (efm) (Use the same procedure used to calculate molar mass.) 2. Divide the known molar mass by the efm 3. Multiply the formula subscripts by this value to get the molecular formula

Answer Molar mass (efm) is 30.0 g 60.0 g divided by 30.0 g = 2 Answer: C2H8N2

Practice Problems 1) What is the empirical formula of a compounds that is 25.9% nitrogen and 74.1% oxygen? 2) Calculate the empirical formula of a compound that is 32.00% C, 42.66% O, 18.67% N, and 6.67% H. 3) Calculate the empirical formula of a compound that is 42.9% C and 57.1% O.

Practice Problems 4) What is the molecular formula for each compound: a) CH2O, 90 g/mol b) HgCl, 472.2 g/mol c) C3H5O2, 146 g/mol