Mass Ratios & Law of Multiple Proportions Miss Fogg Fall 2015.

Slides:



Advertisements
Similar presentations
Empirical and Molecular Formulas
Advertisements

Copyright Sautter EMPIRICAL FORMULAE An empirical formula is the simplest formula for a compound. For example, H 2 O 2 can be reduced to a simpler.
Empirical and Molecular Formulas
Calculate the Empirical Formula for a compound with the following composition: 46.16% carbon; 53.84% nitrogen 1)Change % to grams (if needed) 2)Convert.
Chapter 7 – The Mole and Chemical Composition
Chemistry Notes Empirical & Molecular Formulas. Empirical Formula The empirical formula gives you the lowest, whole number ratio of elements in the compound.
Lecture No. 1 Laws of Chemical Combinations Chemistry.
Chapter 5: Atomic Theory: The Nuclear Model of the Atom
Percent Composition, Empirical Formulas, Molecular Formulas.
Law of Definite Composition and Law of Multiple Proportions
Percent Composition, Empirical Formulas, Molecular Formulas
Mass Conservation in Chemical Reactions Mass and atoms are conserved in every chemical reaction. Molecules, formula units, moles and volumes are not always.
Warm-Up: To be turned in 3.6 mol NaNO 3 = ______ g g MgCl 2 = ______ mol.
Chemistry There are some compounds that we know have elements in fixed mass proportions. Water H 2 O  2 g : 16 g Carbon dioxide CO 2 
Sec. 10.4: Empirical & Molecular Formulas
Empirical Formulas 4/1 - ATB: If an unknown sample has 1 mole of H and 1 mole of Cl, what would be the formula of this compound?
The Mole and Chemical Composition
History of the Atomic Theory. Law of Definite Proportions A given compound contains the same elements in exactly the same proportions by mass, regardless.
Mole Relationships. By definition: 1 atom 12 C “weighs” 12 amu Atomic mass is the mass of an atom in atomic mass units (amu) The mole (mol) is the amount.
The Mole and Chemical Composition
The Mole AA. Review Must turn in your packet with notes stapled to it before you can take the test.
Unit 10 – The Mole Essential Questions:
Empirical and Molecular formulas. Empirical – lowest whole number ratio of elements in a compound Molecular – some multiple of the empirical formula Examples:
Determining Chemical Formulas
Using the MOLE. Percentage composition Percentage composition is the mass of individual elements in a compound expressed as a percentage of the mass of.
Law of Definite Proportions and Law of Multiple Proportions
Empirical and Molecular Formulas.  Tells us relative number of atoms of each element it contains  Example: H 2 O: 2 atoms of H per 1 atom of O  ALSO:
The Laws of Chemistry. Dalton's Atomic Theory A. Elements are composed of extremely small particles atoms called atoms. All atoms of the same element.
Atomic Structure PSC Chapter 3. Atomic Theory of Matter Evidence of atoms Law of Definite Proportions Law of Conservation of Mass Law of Multiple Proportions.
Percentage Composition. Percent Composition Compounds are ALWAYS composed of elements in fixed ratios. This is often referred to as the Law of Definite.
Percent Composition, Empirical and Molecular Formulas.
Empirical & Molecular Formulas
6.10 Calculating Empirical Formulas The empirical formula gives the lowest whole- number ratio of the atoms of the elements in a compound. The empirical.
Chapter 8 Chemical Composition Chemistry 101. Atomic mass unit (amu) = × g Atomic Weight Atoms are so tiny. We use a new unit of mass:
Determining Chemical Formulas
Percent by Mass Miss Fogg Fall 2015.
Today we will:. Learn about Elements and Compounds.
THURSDAY, 19 TH SEPTEMBER 2013 Today in Chemistry Percent Composition Assignment Read pp
Law of conservation of mass states that mass can neither be created nor destroyed in a chemical reaction.
Empirical Formulas. Gives the lowest whole-number ratio of the elements in a compound. Example: Hydrogen Peroxide (H 2 O 2 ) Empirical Formula- HO.
USING MOLAR CONVERSIONS TO DETERMINE EMPIRICAL AND MOLECULAR FORMULAS.
Emperical and Molecular Formulas. If I have 78.5 L of O 2 gas at STP, how many grams would that be? How many molecules would be present? How many atoms.
6.7 Empirical Formula. Formulas Percent composition allows you to calculate the simple ratio of the atoms in a compound. Empirical Formula: formula of.
Molecular Weight, Percent Composition, Empirical Formula.
Calculating Empirical Formulas
Empirical Formulae The empirical formula of a compound is the simplest ratio of the different atoms in it. For example, for ethane (C2H6)it is CH3. You.
The Mole Honors Chem. -How do we measure chemical quantities? -What units of measure do we use?
Atomic Unit Calculations. Calculating Atomic Mass Units (amu) Definition: A unit of mass used to express atomic and molecular weights.
Warm Up Name the two kinds of mixtures List three different separation techniques Is iron rusting a chemical or physical change? If 2g of potassium (K)
Percent Composition, Empirical and Molecular Formulas.
EMPIRICAL AND MOLECULAR FORMULA.  Empirical Formula – The lowest whole number ratio of atoms in a compound. Example: The empirical formula for the compound.
Percent Composition. – Compounds are ALWAYS composed of elements in fixed ratios. This is often referred to as the Law of Definite Proportions. – This.
Law of Definite Composition and Law of Multiple Proportions
How do chemists determine the formula of compounds?
Calculating Empirical Formulas
Empirical Formula.
EMPIRICAL FORMULA VS. MOLECULAR FORMULA .
Simplest Chemical formula for a compound
EMPIRICAL FORMULA The empirical formula represents the smallest ratio of atoms present in a compound. The molecular formula gives the total number of atoms.
Law of Definite Composition and Law of Multiple Proportions
Law of Definite Composition and Law of Multiple Proportions
Introduction to Chemical Principles
Empirical and Molecular Formulas
EMPIRICAL FORMULA The empirical formula represents the smallest ratio of atoms present in a compound. The molecular formula gives the total number of atoms.
Ch. 7-3 Percent Composition
Law of definite and multiple proportions
Percentage Composition
Calculating Empirical and Molecular Formulas
Calculating Empirical and Molecular Formulas
Presentation transcript:

Mass Ratios & Law of Multiple Proportions Miss Fogg Fall 2015

Mass Ratios Atomic mass = mass of an atom measured in amu (atomic mass units) Mass of oxygen =  16.00amu Mass of hydrogen =  1.01amu

Mass ratios Mass of element X : Mass of element Y How to calculate mass ratios: ◦ 1) Find the mass of each element ◦ 2) Divide each elemental mass by the smallest element mass ◦ 3) Express as ratio and round

Example: H 2 O Find the ratio of oxygen to hydrogen in H 2 O ◦ 1) Find the mass of each element O: 16.00amu x 1 atom = 16.00amu H: 1.01amu x 2 atom =2.02amu

Find the ratio O : H ◦ 2) Divide each elemental mass by the smallest element mass O: 16.00amu ÷ 2.02 = 8.0 H: 2.02amu ÷ 2.02 = 1.0 Example: H 2 O

Find the ratio O : H ◦ 3) Express as ratio and round O: 8.0  8 H: 1.0  1 (Law of definite proportions) In H 2 O, oxygen and hydrogen always have a ratio of “16 to 2” or “8 to 1” by mass 8 : 1 Example: H 2 O

Example: KCl KCl always contains one atom of K for every one atom of Cl Find the ratio K : Cl ◦ 1) Find the mass of each element K: 39.10amu x 1 atom = 39.10amu Cl: 35.45amu x 1 atom =35.45amu

Example: KCl Find the ratio K : Cl ◦ 2) Divide each elemental mass by the smallest element mass K: 39.10amu ÷ = Cl: 35.45amu ÷ 35.45= 1.000

Example: KCl Find the ratio K : Cl ◦ 3) Express as ratio and round K:  1.1 Cl:  1 (Law of definite proportions) In KCl, potassium and chlorine always have a ratio of “39.09 to 35.45” or “1.1 to 1” by mass. 1.1 : 1

Law of Multiple Proportions Dalton’s Law (1804): When 2 elements form different compounds, the mass ratio of the elements in one compound is related to the mass ratio in the other by a small whole number

Nitrogen & Oxygen You have more than one molecule made up of the same elements N 2 Onitrous oxide NOnitric oxide NO 2 nitrogen dioxide OO O ON N N N

Nitrogen & Oxygen N 2 Onitrous oxide Mass nitrogen: 14.01amux 2 = 28.02amu Mass oxygen: 16.00amux1= 16.00amu ONN

Nitrogen & Oxygen N 2 Onitrous oxide Mass nitrogen: 28.02amu Mass oxygen: 16.00amu How many grams of N are used every 1g of O? 28.02amu nitrogen= ? 16.00amu oxygen 1g oxygen ONN 1.751g nitrogen

Nitrogen & Oxygen Mass of N for 1g O N 2 Onitrous oxide1.75g NOnitric oxide NO 2 nitrogen dioxide OO O ON N N N

Nitrogen & Oxygen NOnitric oxide Mass nitrogen: 14.01amux 1 = 14.01amu Mass oxygen: 16.00amux1= 16.00amu ON

Nitrogen & Oxygen NOnitric oxide Mass nitrogen: 14.01amu Mass oxygen: 16.00amu How many grams of N are used every 1g of O? 14.01amu nitrogen= ? 16.00amu oxygen 1g oxygen ON g nitrogen

Nitrogen & Oxygen Mass of N for 1g O N 2 Onitrous oxide1.75g NOnitric oxide0.876g NO 2 nitrogen dioxide OO O ON N N N

Nitrogen & Oxygen NO 2 nitrogen dioxide Mass nitrogen: 14.01amux 1 = 14.01amu Mass oxygen: 16.00amux 2= 32.00amu ONO

Nitrogen & Oxygen NO 2 nitrogen dioxide Mass nitrogen: 14.01amu Mass oxygen: 32.00amu How many grams of N are used every 1g of O? 14.01amu nitrogen= ? 32.00amu oxygen 1g oxygen ON g nitrogen O

Nitrogen & Oxygen Mass of N for 1g O N 2 Onitrous oxide1.75g NOnitric oxide0.876g NO 2 nitrogen dioxide0.438g OO O ON N N N How are all these numbers related?

Nitrogen & Oxygen Mass of N for 1g O N 2 Onitrous oxide1.75g NOnitric oxide0.876g NO 2 nitrogen dioxide0.438g OO O ON N N N How are all these numbers related? Law of multiple proportions! 1.75 ÷ = = ÷ = ÷ = = 4

Ch3 Homework - Due tomorrow (10/7): Assessment Problems Page 76: 20-24; 30 Page 83: Problems 60-69; 74-75