SL Chemistry · •The empirical formula of a compound is the simplest whole-number ratio of atoms...

SL ChemistryFriday August 28th 2015

Monday, August 31, 15

Agenda

Warm Up: NONE

Empirical and Molecular Formula Notes

E & M handout (HW)

Internal Assessment - Project Guides


HOMEWORK

Topic 1.1 Introduction to Particulate Nature of Matter & Chemical Change vocabulary

DUE September 4th 2015

The Mole Concept

DUE September 2nd 2015

Empirical & Molecular Formulas



Electronic Copies

I have access to electronic copies of the chemistry textbook and the course companion study guide... PLEASE give me your flash drive and I will give you both copies!

You are more than welcome to have both the electronic and hard copy of the textbook if you prefer though


Stoichiometric Relationships

Introduction to the Particulate Nature of Matter and Chemical ChangeThe Mole Concept

Reacting Masses and Volumes

Ms. Thompson - SL ChemistryWooster High School


Topic 1.2The Mole Concept

• The mole is a fixed number of particles and refers to the amount, n, of substance.

• Masses of atoms are compared on a scale relative to 12C and are expressed as relative atomic mass (Ar) and relative formula/molecular mass (Mr).

• Molar mass (M) has the units g mol-1.

• The empirical formula and molecular formula of a compound give the simplest ratio and the actual number of atoms present in a molecule respectively.


The Mole Concept

Experimental empirical and molecular formula determination

• Empirical describes information that is derived through observation and/or investigation.• Qualitative analysis focuses on determining which elements are present in a compound. It could also verify the purity of the substance.• Quantitative analysis enables chemists to determine the relative masses of elements which allows them to work out their exact composition.• The molecular formula is the actual number of atoms or amount of elements in one structural unit or one mole of the compound. •The empirical formula of a compound is the simplest whole-number ratio of atoms or amount (in mol) of each element present in a compound.

Substance Molecular Formula Empirical FormulaHydrogen Peroxide H2O2 HO


ClassworkWork with a partner and identify the substances

molecular formula and empirical formula

Substance Molecular Formula Empirical Formula

ethane C2H6

water H2O

butanoic acid H2O2

glucose C4H8O2

naphthalene C10H8

benzene C6H12O6


ClassworkWork with a partner and identify the substances

molecular formula and empirical formula

Substance Molecular Formula Empirical Formula

ethane C2H6 CH3

water H2O H2O

hydrogen peroxide H2O2 HO

glucose C4H8O2 C2H4O

naphthalene C10H8 C5H4

benzene C6H12O6 CH2O


The Mole Concept

Empirical Formulas

• When a compound’s percent composition is known, its formula can be calculated.

• First, determine the smallest whole-number ratio of the moles of elements in the compound.

• This ratio gives the subscripts in the empirical formula.• The empirical formula might or might not be the same as the

actual molecular formula! • The empirical formula = molecular formula for IONIC

COMPOUNDS - ALWAYS!


The Mole Concept

Calculating Empirical Formulas

Use the following poem to remember the steps:

Percent to massMass to molesDivide by small

Multiply ‘til whole

EXAMPLE'PROBLEM'

•  Methyl'acetate'is'a'solvent'commonly'used'in'some'paints,'inks,'and'adhesives.'Determine'the'empirical'formula'for'methyl'acetate,'which'has'the'following'chemical'analysis:'48.64%'carbon,'8.16%'hydrogen,'and'43.20%'oxygen.'


Practice Problem

Methyl acetate is a solvent commonly used in some paints, inks, and adhesives. Determine the empirical formula for methyl acetate, which has the following chemical analysis: 48.64% carbon, 8.16% hydrogen,

and 43.20% oxygen.


Step One: % to g Let’s assume we have a 100.0 g sample of methyl acetate. This means that each element’s percent is

also the number of grams of that element.

48.64% C = 48.64 g C8.16% H = 8.16 g H

43.20% O = 43.20 g O


Step Two: g to molConvert each mass into moles using the molar mass of each

element.

48.64 g C x 1 mol C = 4.053 mol C 12.0 g C8.16 g H x 1 mol H = 8.08 mol H 1.01 g H43.20 g O x 1 mol O = 2.700 mol O 16.0 g O


Step Three: ÷ by smallOxygen accounts for the smallest number of moles in the formula, so

divide each element by oxygen’s number of moles: 2.700 mol

Carbon: 4.053 mol / 2.700 mol = 1.501 = 1.5Hydrogen: 8.08 mol / 2.700 mol = 2.99 = 3Oxygen: 2.700 mol / 2.700 mol = 1.000 = 1

Remember, we will want whole-number ratios...


Step Four: x til wholeIn the previous slide, the ratio of C:H:O is 1.5:3:1We need a whole-number ratio, so we can multiply everything by

2 to get rid of the 1.5

C –>3, H –>6, O –>2 So, the final empirical formula is C3H6O2


The Mole Concept

Calculating Molecular Formulas

Sorry, no silly poem here! ☹Step 1 – Calculate the empirical formula (if needed)Step 2 – GIVEN molecular mass (experimental)/empirical formula

molar mass = multiplierStep 3 – Multiply the empirical formula subscripts by the multiplier

found in Step 2

EXAMPLE'PROBLEM'

•  Methyl'acetate'is'a'solvent'commonly'used'in'some'paints,'inks,'and'adhesives.'Determine'the'empirical'formula'for'methyl'acetate,'which'has'the'following'chemical'analysis:'48.64%'carbon,'8.16%'hydrogen,'and'43.20%'oxygen.'


Practice Problem

Succinic acid is a substance produced by lichens. Chemical analysis indicates it is composed of 40.68% carbon, 5.08% hydrogen, and 52.24% oxygen and has a molar mass of 118.1 g mol-1. Determine the empirical

and molecular formulas for succinic acid.


Step One: Find Empirical Formula40.68 % C = 40.68 g C x 1mol C = 3.390 mol C 12.0 g C5.08 % H = 5.08 g H x 1 mol H = 5.03 mol H 1.01 g H54.24% O = 54.24 g O x 1 mol O = 3.390 mol O 16.0 g OC: H: O3.390 mol: 5.03 mol: 3.390 mol3.390 mol 3.390 mol 3.390 mol1 : 1.48: 1 –> 1 : 1.5 : 1, multiply by 2 –> C2H3O2

Round up! Obtain a

WHOLE

number


Step Two: Divide Molar Masses

Molar mass empirical formula = (2 x 12.0 g/mol) + (3 x 1.01 g/mol) + (2 x 16.0 g/mol) = 59.0 g/mol

Given molar mass = 118.1 g/mol

Multiplier = 118.1 g/mol = 2 59.0 g/mol


Step Three: Use Multiplier

Empirical Formula = C2H3O2

x 2 from step two

Molecular formula = C4H6O4


Determination of the Empirical Formula of Magnesium Oxide

You will have a lab next Friday (September 4th)• Your job is to write a thorough and complete exploration

section for this lab.• To achieve this you must:• Research the scientific content of this topic• Write and define a workable research question• Adapt or design a methodology (how you will carry

out your investigation)• Practice academic integrity and cite any sources you

use for your research• Must include safety, environmental, and ethical

considerations, if applicable.Monday, August 31, 15

Determination of the Empirical Formula of Magnesium Oxide

You will have a lab next Friday (September 4th) • Requirements:

• Typed, single-spaced, 12-point Times New Roman font• Include cover page• List and identify your variables:

• Independent variables - ones you manipulate or change• Dependent variables - ones that respond to the change• Controlled variables - ones you do not change and hold constant throughout the experiment

• List of materials you will use (diagrams can be helpful to explain experimental set up)• Methodology - the steps necessary to complete the experiment without you having to be there to explain it. BE CLEAR, CONCISE, AND TO THE POINT.


HOMEWORK

Topic 1.1 Introduction to Particulate Nature of Matter & Chemical Change vocabulary


The Mole Concept

DUE September 2nd 2015

Empirical & Molecular Formulas



SL Chemistry · •The empirical formula of a compound is the simplest whole-number ratio of atoms...

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