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## Chemistry 111 notes

by: Jamisha Evans

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# Chemistry 111 notes CHEM 111 003

Jamisha Evans
EKU
GPA 3.71

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These notes cover chapters 3.5 and 3.7
COURSE
General Chemistry 1
PROF.
Dr. James Patton
TYPE
Class Notes
PAGES
2
WORDS
CONCEPTS
Chem, 111
KARMA
25 ?

## Popular in Chemistry

This 2 page Class Notes was uploaded by Jamisha Evans on Friday September 30, 2016. The Class Notes belongs to CHEM 111 003 at Eastern Kentucky University taught by Dr. James Patton in Fall 2016. Since its upload, it has received 5 views. For similar materials see General Chemistry 1 in Chemistry at Eastern Kentucky University.

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Date Created: 09/30/16
Chemistry 111 Lecture Notes (Dr. Patton) 3.5 Percent Composition of compounds  Percent Composition  Definition: The percentage by mass contributed by each element in the substance  Formula: %element= (n)(molar mass of element)/(molar mass of the compound) x 100  n: Integer, subscript/moles of a particular element EXAMPLE: Percent composition of sulfuric acid (H S2 ) 4 1. Calculate the molar mass of the compound 2 H: 2.016g Sum 4 O: 64.00g of this is 1 S: 32.07g 98.08g 2. Calculate the percent by mass of each element in formula… I will only show how to calculate the percent by mass for hydrogen but the same steps are followed for oxygen and sulfur. %H= (2)(1.008)/(98.09) x 100= 2.055% **The sum of all the percentages should equal 100%**  The percent composition by mass is the same for the molecular formula and the empirical formula of a compound. This means that the empirical formula can be used to calculate the empirical formula of a compound  Empirical formula from percent by mass composition  EXAMPLE: calculate the empirical formula of this unknown compound using percent composition by mass composition. Given C:10.4% S:27.8% Cl:61.7% Steps 1. Convert each of our data into grams assuming we have 100g of sample C:10.4g S:27.8g Cl:61.7g 2. Determine number of moles by using molar mass as a conversion factor C:10.4gC x 1molC/12.01gC=.866 mol C S:27.8gS x 1 molS/32.06 g S= .866 mol S Cl:61.7gCl x 1 molCl/34.45g Cl = 1.74 mol Cl 3. Find whole number containing ratio by dividing by the smallest amt of mol C:.866/.866 =1 S:.866/.866= 1 Cl:1.74/.866= 2 4. The least electronegative element is listed first and the most electronegative is listed last. (more about this is Ch. 8) Final answer: CSCl 2  Molecular formula from empirical formula EXAMPLE: Empirical formula for a compound is CH and i3s molar mass is 30.1 g/mol. What is the molecular formula of the unknown?  Use this formula  n= (molar mass of compound)/(empirical formula’s molar mass) n= (30.1g)/(15.03) **15.03 is the mass of CH 3* n=2 …Now multiply each subscript by 2 C 1 x 2= 3 x 6= Final answer:C 2 6  Mass of element from percent by mass composition EXAMPLE: Mass of aluminum (grams) in 371g of aluminum oxide? Mass of Al 2 3 moles of Al2O 3 moles Al mass of Al Final answer: 196 g Al 3.7 Chemical Reactions and Chemical Equations  Chemical reaction  Definition: process in which a substance or substances are changed into one or more new substances  Chemical equation  Definition: Uses chemical symbols to represent this process  Chemical equations must be balanced  The physical state of all the reactants and products are represented  Gas: (g)  Liquid: (l)  Solid: (s)  Aqueous: (aq)  Make sure to balance the equation. Change only the coefficient NOT the subscripts EXAMPLE: write a chemical equation for the formation of ammonia from hydrogen and nitrogen gas 3H 2g) + N 2g) 2NH 3(g) Product Reactant H:6 H:6 N:2 N:2

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