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Chemistry (28)
CHEM 1F92 (17)
Chapter 3

# Chapter 3.docx

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School
Brock University
Department
Chemistry
Course
CHEM 1F92
Professor
Lydia W.L.Chen
Semester
Fall

Description
Chapter 3 – Mass Relationships in Chemical Reactions 3.1 –Atomic Mass • Atomic mass is the mass of an atom, measured in atomic mass units (amu). o One atomic mass unit (amu) is a mass equal to 1/12 the mass of a carbon-12 atom. • Once the atomic mass of carbon-12 was set to 12 amu, all other elements were given amu relative to carbon-12. o For example, through experiments, it was determined that the mass of an H atom is 8.400% the mass of a carbon-12 atom (1.008 amu). Average Atomic Mass • Most of the elements in nature have more than one isotope. • This means that when calculating the atomic mass of an element, we must find the average atomic mass of the different isotopes that exist in nature. 12 13 o C is 98.90% abundant (12.00 amu) and C is 1.10% abundant (13.00335amu). Average Atomic Massof NaturalCarbon=(0.9890)(12.00000amu)+(0.0110)(13.00335amu) ¿12.01amu • This value is slightly over 12.00 amu since a carbon-13 isotope also occurs in nature, but in small amounts. • In the periodic table, average atomic masses appear for all the elements. However, for simplicity, we refer to them as atomic mass. 3.2 –Avogadro’s Number and theAtomic Mass of an Element • The mole is the amount of a substance that contains as many elementary entities (atoms, molecules, ions, and formula units) as there are atoms in 12 grams of carbon-12. o The number of atoms in 12 g of carbon is known as Avogadro’s number (N ). A N A 6.022 × 10 23 • Molar mass (M) is the mass per mole of units (atoms, molecules....) of a substance (in grams/mole). • The molar mass of an element is the same as the atomic mass of the substance. 1amu=1 g mol 1g 1amu= 23 6.02×10 units 1amu=1.66×10 −2grams 1g=6.02×10 amu3 Look at examples 3.2, 3.3, and 3.4 3.3 – Molecular Mass • The molecular mass is the sum of the atomic masses of all the atoms in a molecule. o It is measured in amu (Not g/mol. Only molar mass is measured in g/mol). Molecular Mass of H O = 2 (atomic mass of H) + 1(atomic mass of O) 2 = 2(1.008 amu) + 1(16.00 amu) = 18.02 amu • For ionic compounds such as NaCl and MgO, the term formula mass is used instead of molecular mass, since these are formula units. Formula Mass of NaCl = (atomic mass of Na ) + 1(atomic mass of Cl )- = (22.99 amu) + (35.45 amu) = 58.44 amu 3.4 – The Mass Spectrometer • In a mass spectrometer, a gaseous sample of an element is bombarded into a stream of high energy electrons. • When the gaseous atoms/molecules collide with these electrons, they each lose an electron, and become a positively charged ion. • These ions pass through oppositely charged plates and accelerate as they pass through them. • After leaving the plates, they are deflected by a magnet and begin to curve. The amount of curve depends on the charge-to-mass ratio of each of the ions. • Since the charges of the ions are all the same, their differences in mass would cause different deflections. Higher masses would deflect less, whereas lower masses would deflect more. • The atoms/molecules that had similar masses would strike the detecting screen around the same location (each ion would create a small current as it strikes the screen). Using this info, the mass of the different isotopes and their abundance could be calculated. • The first mass, spectrometer, discovered in the 1920s by F. W.Aston, provided evidence that Neon isotopes do exist (neon-20 and neon-22). Neon-21 was discovered later using a more accurate mass spectrometer. 3.5 – Percent Composition of Compounds • Percent composition by mass is the percentage (by mass) of each element present in a compound. (n)(M ) PercentComposition()= element(100 ) ( M compound) • n represents the amount of moles of the element for every mole of the compound (subscript to element) • In the formula, Mcompoundn also be replaced with empirical formula o This is because when the empirical formula is used, both the n and M compoundlues would decrease (by the same factor), leading to the same % composition Refer to Example 3.8 – 3.10 When we are given the % composition and have to determine the empirical formula, assume 100 g of sample, and find the mass of each element in the sample. Then determine moles, mole ratios of the elements and then the empirical formula 3.6 – Experimental Determination of Empirical Formulas Consider a reaction in which ethanol is burned in the presence of Oto form H O and CO . 2 (g) 2 2 Since no C or H atoms are present in the2O , all the C and H atoms will be in the ethanol. Although O atoms are presen
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