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Percentage composition Indicates the relative amount of each element present in a compound.
Calculating percentage composition Step 1 : Calculate molar mass Step 2 : Divide the subtotal for each elements mass by the molar mass. Step 3: Multiply by 100 to convert to a percentage.
Example 1 1. Calculate the molar mass of water, H 2 O. Step 1 – H- 2(1.01)= 2.02 O – 1(16.0)= 16.0 ________ 18.02
Step /18.02 X 100 = 11.21% 16.0 /18.02 X 100 = % Water is composed of % hydrogen and % oxygen.
Example 2 Calculate the percentage composition of sucrose, C12H22O11.
Example 3 Find the percentage composition of hydrogen in sulfuric acid.
Percent Composition Percent = part / whole Example: MgO Find molar mass of whole compound: MgO = = 40.3 grams % Mg (by mass) = mass of Mg.
IIIIII Percent Composition Empirical Formula Molecular Formula Chemical Formulas.
Empirical and Molecular Formulas Unit Empirical and Molecular Formulas An empirical formula shows the simplest whole number ratio of atoms of.
Percent Composition The relative amounts of elements in a compound. % composition = mass of element mass of compound x g C 28g CH 4 x 100 = 85.7%
Calculating Empirical Formula. Tells the simplest whole number molar ratio of elements in a compound.
Moles and Formula Mass The Mole 1 dozen = 1 gross = 1 ream = 1 mole = x There are exactly 12 grams of carbon-12 in one mole of.
IIIIII Formula Calculations Ch. 7 – The Mole. A. Percentage Composition n the percentage by mass of each element in a compound EMPIRICAL and MOLECULAR.
IIIIII III. Formula Calculations (p ) Ch. 3 & 7 – The Mole.
IIIIII C. Johannesson III. Formula Calculations The Mole.
Formulas and Percent Composition. Percent Composition The percent composition is the percentage by mass of each element in a compound This helps distinguish.
Definite proportions and percent by mass. Law of definite proportions The Law of Definite Proportions states that a compound is always composed of the.
Lesson Starter Write the stock system names for MnO, PbO 2 Write the stock system names for MnO, PbO 2 Manganese (II) Oxide Manganese (II) Oxide Lead.
Based on lab evidence For example, we know that 4.9 g of magnesium react with 32.0 g of Bromine. How can we figure out the formula of the compound.
Unit One Chemistry – Horsham College Chapter 6 – VCE Chemistry Measurement in Chemistry.
Empirical Formula From percentage to formula. The Empirical Formula The lowest whole number ratio of elements in a compound. The molecular formula the.
Empirical and Molecular Formulas. Definitions Empirical Formula: the lowest whole number ratio of elements in a compound. It can be determined from experimental.
EMPIRICAL FORMULA empirical formulaThe empirical formula represents the smallest ratio of atoms present in a compound. molecular formulaThe molecular formula.
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.
Honors Chemistry Part 1 Unit 4 – The Mole. WHAT IS A MOLE? x
The Mole Chapter 6. The Mole Remember it is a measurement for anything Remember it is a measurement for anything 1 mole means x of anything.
Calculations Aim: to revise key concepts about molar calculations.
Stoichiometry Sherlock Holmes, in Sir Arthur Conan Doyles A Study in Scarlet In solving a problem of this sort, the grand thing is to be able to reason.
Empirical and Molecular Formulas Part 2: Calculations using percent composition.
Use of Formulae Masses and Moles (but no badgers).
It’s just like finding out your test score!!! 97 correct out of 100 questions = 97 x 100 = 97% 100 Total of “something” x 100 = % TOTAL.
Using the methods of stoichiometry, we can measure the amounts of substances involved in chemical reactions and relate them to one another. Stoichiometry.
1 Chapter 7- Molecular Formulas. 2 Warm Up: Determine the name and molar mass of the following compounds. NaOH MnO 2 H 2 SO 4.
Topic 14 Topic 14 Topic 14: The Mole Table of Contents Topic 14 Topic 14 Basic Concepts Additional Concepts.
IIIIIIIV Ch. 3 & 7 – The Mole I. Molar Conversions (p.80-85, )
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