Quantities in Chemistry Stoichiometry Ch 5 Detecting elements and compounds Mass spectrometry is a highly sensitive technique that can be used to help identify compounds. Samples of only a few milligrams are required. Mass spectrometry is useful in the identification of unknown materials, because the mass spectrum of an unknown sample can be compared with a database of known spectra. This makes it useful in forensic science.
Spectra of samples could be compared, for example, to link a suspect to a weapon that has been fired. Mass Spectrometer Currently most accurate way to evaluate the mass of an atom. Atoms or molecules are passed into a beam of high speed electrons which coverts them into positive ions. The positive ions are then accelerated through an electric field and deflected by a magnetic field. The lightest and most highly charged element is deflected the
most. Eg: 63Cu+is deflected more than 65Cu+ Eg: 63Cu2+is deflected more than 63Cu+ A mass spectrum shows information about the relative isotopic mass (Ir), percentage abundance of each isotope and the number of isotopes in a given sample. Mass spectrometry Process of mass spectrometry Process of mass spectrometry
Mass spectra of monatomic elements Relative Atomic Mass (Ar) The relative atomic mass of an element represents the weighted average of the relative isotopic masses of all the naturally occurring isotopes of a sample of an element relative to carbon-12 which is assigned a mass of exactly 12. It takes into consideration the relative abundance of each isotope. As masses are relative to carbon-12 Ar has no units. Relative atomic mass data can be found on your periodic table.
Can be calculated using this formula: Ar = (Ir of 1st isotope x % abundance) + (Ir of 2nd isotope x % abundance)..... 100 Relative Isotopic Mass (Ir) Defined as The mass of an individual isotope of an element on the relative atomic mass scale, on which the masses of particles are compared with the mass of the carbon-12 isotope. All other isotopic masses are measured relative to carbon-12.
Masses of other isotopes and atoms are determined in mass spectrometers. Since masses are relative, relative isotopic mass has no units. Using mass spectra to calculate Ar The mass spectrum of an element indicates the mass and abundance of each isotope present. For example, the mass spectrum of boron indicates two isotopes are present: abundance (%) 100
11 80 B (80%) 60 40 10 B (20%)
20 0 0 2 4 6 m/z 8
10 12 How can this be used to calculate the Ar of boron? Calculating Ar Most elements have more than one isotope. The relative atomic mass of the element is the average mass of the isotopes taking into account the abundance of each isotope. Example: what is the Ar of boron? In a sample of boron, 20% of the atoms are 10B and 80% are 11B.
If there are 100 atoms, then 20 atoms would be 10B and 80 atoms would be 11B The relative atomic mass is calculated as follows: Ar of B = (20 10) + (80 11) 100 Ar of B = 10.8 Counting atoms and molecules When conducting a chemical reaction, it is often important to mix reactants in the correct proportions. This prevents contamination of the products by wasted reactants. However, atoms are very
small and impossible to count out. In order to estimate the number of atoms in a sample of an element, it is necessary to find their mass. The mass of an atom is quantified in terms of relative atomic mass. Relative atomic mass The relative atomic mass (Ar) of an element is the mass of one of its atoms relative to 1/12 the mass of one atom of carbon-12.
relative atomic mass average mass of an atom 12 = (Ar) mass of one atom of carbon-12 Most elements have more than one isotope. The Ar of the element is the average mass of the isotopes, taking into account the abundance of each isotope. This is why the Ar of an element is frequently not a whole number.
Relative molecular mass and Relative formula mass (Mr ) Relative molecular mass: the mass of a molecule relative to carbon-12. Applies only to molecular substances. Relative formula mass: the sum of the relative atomic masses of the atoms as given in the chemical formula of a species. Applies to molecular and nonmolecular substances such as ionic solids. Calculated by adding up the atomic masses of the atoms as given by the formula of the substance. Eg:Mr(H2O) = 2 x Ar(H)+ Ar(O) = 2 x 1.0 + 16.0 = 18.0
Relative molecular mass The relative molecular mass (Mr) of a covalent substance is the mass of one molecule relative to 1/12 the mass of one atom of carbon-12. Mr can be calculated by adding together the masses of each of the atoms in a molecule. Example: what is the Mr of H2SO4? 1. Count number of atoms (2 H) + (1 S) + (4 O) 2. Substitute the Ar values
(2 1.0) + (1 32.1) + (4 16.0) 3. Add the values together 2.0 + 32.1 + 64.0 = 98.1 Relative formula mass The equivalent of relative molecular mass for an ionic substance is the relative formula mass. This is the mass of a formula unit relative to 1/12 the mass of one atom of carbon-12. It is calculated in the same way as relative molecular mass, and is represented by the same
symbol, Mr. Example: what is the Mr of CaCl2? 1. Count number of atoms (1 Ca) + (2 Cl) 2. Substitute the Ar values (1 40.1) + (2 35.5) 3. Add the values together 40.1 + 71.0 = 111.1
What is the mole? The mole (n) Atoms are extremely tiny, so chemists never deal with them one at a time. The mole is the amount of substance that contains as many particles ( atoms, ions or molecules) as there are atoms in exactly 12g of the carbon-12 isotope. The number of atoms has been experimentally estimated to be 6.02 x 1023 (Avogadros constant L). To find the mass of one mole of an element add
g to the relative atomic mass of that element. Moles and Avogadro's number Converting number of particles to number of moles and vice versa Number of moles (n)= Number of particles (N) Avogadros constant (NA) Number of particles(N)=number of moles(n)x Avogadros(NA) N n
NA What is the mole? A mole is the amount of a substance of a system which contains as many elementary particles as there are Carbon atoms in 12 grams of 12C STANDARD: 12 grams of 12C One Mole of 12C MICROSCOPIC LEVEL: Number of particles
X of 12C atoms X=6.02 x 10 23=Avogadros number MACROSCOPIC LEVEL: Mass 12 grams WHY 12C and 12 g? As long as they have the same number of atoms, the mass ratios between elements do not change. Therefore, we can count the number of particles by weighing. 1 mole = the number of
12 C atoms in 12 g of 12C ATOMIC LEVEL MACROSCOPIC LEVEL Molar mass (a mass per mole) Relative Atomic/Molecular Mass Proportional relationships relative to C-12
C 12 C 12 g O 16 O
16 g H 1 H 1g N
14 N 14 g Proportional relationships relative to C-12 mm 3 grams
The mass ratio between the same number of the sweets 1 6 grams 2 The mass per sweet
m 0.3 g 10 mm 10 jelly beans Jelly Bean 0.6 g The mass per sweet
mm m X mm 0.3 g The mass of X sweets 800 grams X jelly beans Jelly bean
0.6 g ? Relative Atomic Mass / Molecular Mass C 12
Molar Mass ( a mass per mole) 1 MOLE 1 MOLE O 16 C 12 g
O 16g Then... 12g of C has the same number of atoms as 14g of N. 24g of C has the same number of atoms as 28g of N. 32g of O has twice the number of particles than 12g of C.
Molar mass (M) The mass of 1 mole of the element. The unit is grams per mole (g/mol) Molar mass of carbon atoms = mass of 1 mol of C atoms = 12 g/mol 12g of carbon contains 6.02 x 1023 atoms of carbon Can be worked out by calculating the relative molecular mass (Mr) of a substance and use the unit g/mol. Molar mass (M) Molecular compounds:
Numerically equal to the relative molecular mass (Mr ) expressed in grams per mole Molar mass of water molecules = mass of 1 mol of H2O molecules = 18 g/mol 18g of water contains 6.02 x 1023 molecules of water Molar mass (M) Ionic compounds: The relative formula mass is found by adding together the relative atomic mass of each atom in the formula of the compound Mr of CuSO4 = Ar (Cu)+ Ar (S)+ 4x Ar (O)
= 63.5+32.1+4x16 = molar mass 159.6 g/mol Moles, mass and Ar / Mr Converting mass to number of moles and vice versa Number of moles (n) = Mass (m in g) Molar mass (M in g/mol) Mass(m in g)=Number of moles(n)xMolar mass(M in g/mol) m n
M The number aspect 6.02 x1023 1 atom /molecule Micro Relative Atomic/ Molecular Mass 1 mol=6.02x1023
atoms/molecules (NA) 12 grams of C-12 has 6.02x1023 (1mol) atoms 1. Relative mass concept. 2. Mole concept N NA
Macro m = n =N M NA Amount of substance (n) m M Molar Mass
(M) The mass aspect Moles, mass and Mr calculations Formulae calculations Percentage by mass Elemental analysis is an analytical technique used to determine the percentage by mass of certain elements present in a compound. To work out the empirical
formula, the total mass of the compound is assumed to be 100 g, and each percentage is turned into a mass in grams. If necessary, the mass of any elements not given by elemental analysis is calculated. The empirical formula of the compound can then be calculated as normal. Percentage Composition by Mass of a Compound Indicates what proportion of the total mass of the compound is accounted for by the mass of
each element. Can be determined by: Using experimental results involving masses Using the chemical formula Using experimental results involving masses % Element = m (element) m (compound) x 100 3.425g of a sample of an oxide of lead was completely reduced to form 3.105g of lead.
What is the percentage composition of the oxide? Mass of lead oxide: 3.425g Mass of lead: 3.105g Mass of oxygen: 3.425-3.105 = 0.320g %(Pb) = m(Pb) x 100 m ( lead oxide) = 3.105g x 100 3.425g = 90.66% %(O) = m(O)
x 100 m ( lead oxide) = 0.320g x 100 3.425g = 9. 34% Using the chemical formula Types of formulae The empirical formula of a compound shows the relative numbers of atoms of each element present, using the smallest whole numbers of atoms. For example, the empirical formula of hydrogen peroxide
is HO the ratio of hydrogen to oxygen is 1:1. The molecular formula of a compound gives the actual numbers of atoms of each element in a molecule. The molecular formula of hydrogen peroxide is H2O2 there are two atoms of hydrogen and two atoms of oxygen in each molecule. Determining empirical formulae Calculating empirical formulae Empirical Formula Formula that gives the simplest whole number
ratio of number of moles of each atom (or ions for ionic species) in a pure substance. When the mass of each element is known, then the number of moles of each element can be calculated and the ratio of the number of moles and the empirical formula. Empirical Formula Calculating molecular formulae The molecular formula can be found by dividing the Mr by the relative mass of the empirical formula. Example: What is the molecular formula of hydrogen
peroxide given that its empirical formula is HO and the Mr is 34? 1. Determine relative mass of empirical formula: empirical formula mass = H + O = 1.0 + 16.0 = 17 2. Divide Mr by mass of empirical formula to get a multiple: relative molecular mass = 34 = 2 multiple = 17 mass of empirical formula 3. Multiply empirical formula by multiple: HO 2 = H2O2 Molecular Formula
Gives the ACTUAL number of atoms present in a molecule. (Only used for covalent molecular species). Sometimes the empirical formula and the molecular formula are the same. Can be found using the empirical formula if the relative molecular mass (molar mass) of the compound is known. The molar mass allows for a comparison to be made against the empirical formula to determine the whole number multiple.
Molecular Formula To help you find the molecular formula you must find the relative empirical formula mass EFr This is found by adding together the relative atomic masses of the atoms in the empirical formula. C5H4 = C10 H8 What are the different types of yield? The percentage yield of a chemical reaction shows how much product was actually made compared with the amount of product that was expected.
To calculate the percentage yield, the theoretical yield and the actual yield must be calculated. The theoretical yield is the maximum mass of product expected from the reaction, calculated using reacting masses. The actual yield is the mass of product that is actually obtained from the real chemical reaction. Calculating yield The percentage yield of a reaction can be calculated using the following equation:
percentage yield = (actual yield 100) / theoretical yield Example: What is the percentage yield of a reaction where the theoretical yield was 75 kg but the actual yield was 68 kg? percentage yield = (actual yield 100) / theoretical yield = (68 100) / 75 = 90.7%
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