III. Formula Calculations (p )

Slides:



Advertisements
Similar presentations
Ch. 7 – The Mole Formula Calculations.
Advertisements

Percent Composition Empirical Formula Molecular Formula
III. Formula Calculations
III. Formula Calculations (p )
Chapter 11 Empirical and Molecular Formulas
Empirical and Molecular Formulas Empirical and Molecular Formulas An empirical formula shows the simplest whole number ratio of atoms of each element.
Determining Chemical Formulas Experimentally % composition, empirical and molecular formula.
IIIIIIIV Topic 6 The Mole I. Molar Conversions A. What is the Mole? n A counting number (like a dozen) n Avogadro’s number (N A ) n 1 mol = 6.02  10.
Empirical Formula The empirical formula indicates the ratio of the atoms of an element in a compound.
Warm-Up: To be turned in 3.6 mol NaNO 3 = ______ g g MgCl 2 = ______ mol.
1 Empirical Formulas Honors Chemistry. 2 Formulas The empirical formula for C 3 H 15 N 3 is CH 5 N. The empirical formula for C 3 H 15 N 3 is CH 5 N.
Chapter 10 II. Formula Calculations
IIIIIIIV C. Johannesson The Mole I. Molar Conversions.
IIIIIIIV The Mole I. Molar Conversions What is the Mole? A counting number (like a dozen or a pair) Avogadro’s number 6.02  mole = 6.02  10.
IIIIII II. Formula Calculations Ch. 10 – The Mole.
Tis the season to be thankful so lets thank Avogadro for math in Chemistry.
IIIIII C. Johannesson III. Formula Calculations (p ) Ch. 3 & 7 – The Mole.
IIIIII Suggested Reading: Pages Chapter 7 sec 3 & 4.
Percent Composition Like all percents: Part x 100 % whole Find the mass of each component, divide by the total mass.
What Could It Be? Empirical Formulas The empirical formula is the simplest whole number ratio of the atoms of each element in a compound. Note: it is.
IIIIIIIV Chapter 10 – Chemical Quantities What is the Mole? n A unit of measurement used in chemistry. n A counting number like – a dozen eggs, a ream.
IIIIII III. Formula Calculations (p ) Ch. 3 & 7 – The Mole.
IIIIII Formula’s The Mole. Formulas n molecular formula = (empirical formula) n n molecular formula = C 6 H 6 n empirical formula = CH Empirical formula:
IIIIII Formula Calculations The Mole. A. Percentage Composition n the percentage by mass of each element in a compound.
Back side of blue notes Top left section Percent composition by mass: the percentage of an element in a compound How: Find molar mass first (mass of element.
(4.6/4.7) Empirical and Molecular Formulas SCH 3U.
% Composition, Empirical Formulas, & Molecular Formulas Chapter 9 sections 3 & 4.
Calculating Empirical Formulas
C. Johannesson Ch. 11 – The Mole Molar Conversions & Calculations.
IIIIII Molecular Formula Notes The Mole. C. Molecular Formula n “True Formula” - the actual number of atoms in a compound CH 3 C2H6C2H6 empirical formula.
IIIIII II. % Composition and Formula Calculations Ch. 3 – The Mole.
Percent Composition Determine the mass percentage of each element in the compound. Determine the mass percentage of each element in the compound. Mass.
IIIIII Using The Mole. A. Percentage Composition n the percentage by mass of each element in a compound.
The Mole By Mr. M.
Empirical Formula: Smallest ratio of atoms of all elements in a compound Molecular Formula: Actual numbers of atoms of each element in a compound Determined.
Warm-up March 22nd 9.67 moles of KCl = __________________ grams KCl
Writing Empirical and Molecular Formulas
III. Formula Calculations (p )
Empirical Formula.
THE MOLE Formula Calculations.
Percentage Composition
Molar Conversions & Calculations
Empirical and Molecular Formulas
III. Formula Calculations (p )
Formula Calculations (p )
The Mole Formula Calculations.
Molar Conversions (p.80-85, )
III. Formula Calculations
Molar Mass.
EMPIRICAL FORMULA The empirical formula represents the smallest ratio of atoms present in a compound. The molecular formula gives the total number of atoms.
III. Formula Calculations (p )
Ch. 8 – The Mole Empirical formula.
Percent Composition Empirical Formula Molecular Formula
II. Formula Calculations
II. Percent composition
Determining Empirical and Molecular Formulas
III. Formula Calculations
Ch. 8 – The Mole Empirical formula.
II. Percent composition
III. Formula Calculations (p )
% Composition, Empirical Formulas, & Molecular Formulas
Hi Guys & Gals! Warm-Up Time!
% Composition, Empirical Formulas, & Molecular Formulas
Jonathan Seibert PNHS - Chemistry
Empirical and Molecular Formulas
III. Formula Calculations (p )
III. Formula Calculations (p )
Empirical and Molecular Formulae
Ch. 10– The Mole Formula Calculations.
Molecular Formula.
Presentation transcript:

III. Formula Calculations (p. 226-233) Ch. 3 & 7 – The Mole III. Formula Calculations (p. 226-233)

A. Percentage Composition the percentage by mass of each element in a compound

A. Percentage Composition Find the % composition of Cu2S. 127.10 g Cu 159.17 g Cu2S %Cu =  100 = 79.852% Cu 32.07 g S 159.17 g Cu2S %S =  100 = 20.15% S

A. Percentage Composition Find the percentage composition of a sample that is 28 g Fe and 8.0 g O. 28 g 36 g %Fe =  100 = 78% Fe 8.0 g 36 g %O =  100 = 22% O

A. Percentage Composition How many grams of copper are in a 38.0-gram sample of Cu2S? Cu2S is 79.852% Cu (38.0 g Cu2S)(0.79852) = 30.3 g Cu

A. Percentage Composition Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O? 36.04 g 147.02 g %H2O =  100 = 24.51% H2O

C2H6 CH3 B. Empirical Formula Smallest whole number ratio of atoms in a compound C2H6 reduce subscripts CH3

B. Empirical Formula 1. Find mass (or %) of each element. 2. Find moles of each element. 3. Divide moles by the smallest # to find subscripts. 4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

B. Empirical Formula Find the empirical formula for a sample of 25.9% N and 74.1% O. 25.9 g 1 mol 14.01 g = 1.85 mol N = 1 N 1.85 mol 74.1 g 1 mol 16.00 g = 4.63 mol O = 2.5 O

N2O5 N1O2.5 B. Empirical Formula Need to make the subscripts whole numbers  multiply by 2 N2O5

CH3 C2H6 C. Molecular Formula “True Formula” - the actual number of atoms in a compound CH3 empirical formula ? C2H6 molecular formula

C. Molecular Formula 1. Find the empirical formula. 2. Find the empirical formula mass. 3. Divide the molecular mass by the empirical mass. 4. Multiply each subscript by the answer from step 3.

(CH2)2  C2H4 C. Molecular Formula The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is 28.1 g/mol? empirical mass = 14.03 g/mol 28.1 g/mol 14.03 g/mol = 2.00 (CH2)2  C2H4