About Formulas. Types of Bonding Ionic Occurs between metals & nonmetals Electrons are...

About FormulasAbout Formulas

Types of Bonding

Metals vs. Nonmetals

Most of the elements in the periodic table are metals. • Except for H, everything to the left of

the staircase is a metal

Nonmetals are located to the right of the staircase.

Crystal Lattice vs. Molecule Molecule: Discrete unit, definite number

of atoms. All molecules of a given compound are identical. Exact formula.

Crystal Lattice: Regular, repetitive, 3-D arrangement of atoms, ions, or molecules. No set size. Not perfect. No two exactly alike. • Formula gives smallest whole number ratios.

(Formula unit). “Exact” formulas NOT useful.

2-D repetitive patterns

NaCl

Crystals: 3-D repetitive patterns

Formulas

Tell the type & number of atoms in a compound.

Microscopic level: imagine 1 atom, molecule, or formula unit.• Formula gives atom ratios.

Macroscopic level: imagine working in the lab• Formula gives mole ratio

FormulasFormulas

Ex: 2HEx: 2H22O can meanO can mean • 2 molecules of water2 molecules of water• 2 moles of water.2 moles of water.

2H2H22O molecules have 4 hydrogen O molecules have 4 hydrogen atoms & 2 oxygen atoms.atoms & 2 oxygen atoms.

2 2 Moles Moles of Hof H22O molecules have 4 O molecules have 4 moles moles of hydrogen atoms & 2 of hydrogen atoms & 2 molesmoles of oxygen atoms. of oxygen atoms.

Ionic vs. Covalent Formulas By looking at the type of atoms, we can decide if the

substance is an ionic compound or a molecular (covalent) compound.

Molecular compounds usually have all nonmetals.

Ionic compounds usually have metal + nonmetal.

Empirical Formula

Smallest whole number ratio of the elements in the compound.

Ionic compounds have only empirical formulas.

Covalent compounds have empirical and molecular formulas. They can be the same or different.

Molecular Formulas

For covalent (molecular) compounds.

Give the exact composition of the molecule.

Molecular compounds have both empirical and molecular formulas. They can be the same or different.

Say everything you can about:Say everything you can about:

NaClNaCl CC66HH66 CC66HH1212OO66

CHCH44 CFCF44 BeSOBeSO44

CC88HH1818 KIKI CC22HH44ClCl22

CaBrCaBr22 CuSOCuSO445H5H22OO

Ionic, empirical

Covalent, empir., molec?

Covalent, molecular

Ionic, empirical

Covalent, molecular

Covalent, empirical, molecular?

Ionic, empirical

Ionic, empirical

Covalent, molecular

Ionic, empirical

Covalent, molecular

Say everything you can aboutSay everything you can about

PHPH33 CC22HH44 AlAl22OO33

SrISrI22 NFNF33 HH22SeSe

CHCH33OHOH SiOSiO22 HH22OO22

CClCCl44 XeFXeF44 PP44OO1010

Empirical Formulas from % Composition

Convert to mass. Convert to moles. Divide by small. Multiply ‘til whole.

Note 1: last step not always used. Note 2: sometimes you start at step 2, if

they give analysis data in grams.

Compound: 63.52% Fe & 36.48% S

Convert to grams: Assume 100 g sample-size to make math easy.• 63.52 g Fe and 36.48 g S

Convert to moles: Divide by atomic masses.

63.52 g Fe / 55.8 g Fe/mol Fe = 1.138 mol Fe36.48 g S / 32.1 g S/mol S = 1.136 mol S

Compound: 63.52% Fe & 36.48% S

Divide by small: This is where you take the ratio.• 1.136 and 1.138 are essentially the

same. Divide both by 1.136 and get 1 for each.

• Empirical formula is FeS.

26.56 % K, 35.41% Cr, & remainder O

Assume 100-g sample size.• 26.56 g K• 35.41 g Cr• 38.03 g O

Convert to moles.26.56 g K / 39.1 g K/mol K = 0.679 mol K35.41 g Cr / 52.0 g Cr/mol Cr = 0.681 mol Cr38.03 g O / 16.0 g O/mol O = 2.38 mol O

26.56 % K, 35.41% Cr, & remainder O

Divide by small:• 0.679 and 0.681 are essentially the same.• Much smaller than 2.38.• Divide each by 0.68

• K: 1, Cr: 1, O: 3.5 KCrO3.5? No dec’s

Multiply ‘til whole: K2Cr2O7. Multiply all three by 2,

doesn’t change relationship.

Relationship between empirical and molecular formulas

The molecular formula is a whole number multiple of the empirical formula.

Molec. Formula = n (Empirical Form.)

n is a small whole number, which multiplies the subscripts. Sometimes, n = 1.

Molecular Formula

If you know the empirical formula and the molar mass, you can find the molecular formula.

Step 1: Find the mass of the empirical formula.

Step 2: Molar mass Empirical mass = small whole number, n

Step 3: Multiply the subscripts in the empirical formula by n.

Finding Molecular Formulas

Find the molecular formula for the substance whose empirical formula is CH and whose molar mass = 78.0 g

Step 1: Empirical mass = 13.0 g Step 2: Molar mass = n = 78.0/13.0

n = 6. Step 3: 6 X subscripts in CH = C6H6.

Empirical mass

What can you say about CuSO45H2O?

It’s a hydrated salt. For every mole of CuSO4, there are 5 moles

of water. If it’s heated, it dries out. The water goes

into the air and you’re left with the anhydrous salt, CuSO4.

CuSO45H2O CuSO4 + 5H2O

CuSO45H2O CuSO4 + 5H2O

The mole ratio in the formula can be used to predict how much water would be given off by any size sample.• If you had 2 moles of CuSO45H2O, how

much water would you lose on heating?

• If you had 5 moles of CuSO45H2O, how much water would you lose?

Percent Water in CuSO4•5H2O

Step 1: Calculate the formula mass.• Mass of salt + Mass of water

Percent = (Part/Whole) X 100%

% H2O = Mass H2O/Formula Mass X 100 %

% composition from the formula

Percent Water in CuSO4•5H2O

12.00 grams of CuSO4•5H2O are heated gently.

The final mass after repeated heatings is 7.68 grams. What is the % of water?

Mass of H2O = (12.00 – 7.68) = 4.32 g

Percent H2O = 4.32 g X 100% = 36%12.00 g

Percent composition from experimental data

CuSO4xH2O CuSO4 + xH2O

Use experimental data to find x:

12.00 grams of CuSO4•5H2O are heated gently. (Hydrated salt.)

7.68 g = mass anhydrous salt remaining = mass CuSO4.

Mass of H2O = (12.00 – 7.68) = 4.32 g

CuSO4xH2O CuSO4 + xH2O

Note: moles of CuSO4 & moles of H2O can be determined.

Have grams already.• Convert to moles• Divide by small• Multiply ‘til whole.

CuSO4xH2O CuSO4 + xH2O

7.68 g CuSO4 / 159.6 g/mol

= 0.04812 mol CuSO4

4.32 g H2O / 18.0 g/mol

= 0.240 mol H2O Divide by small, which is 0.04812 CuSO45H2O