Chapter 6: Chemical Composition
Molar Mass & Atomic Mass
- (blank) or amu is defined as 1/12 of the mass of a carbon-12 atom -(blank) of any element-- the mass of 1 mol of atoms of that element-- is equal to the atomic mass of that element expressed in amu a) one copper atoms: has (blank0 of 63.55 amu b) 1 mol of copper atoms has a mass of 63.55 g c) (blank) of copper is 63.55 g/mol
More examples of moles
-22 real copper pennies contain about 1 mol of Cu atoms -2 large helium ballons contain approximately 1 mol of He atoms
The conversion factor comes directly from the chemical formula
-8 legs: 1 spider -4 legs: 1 chair -2 H atoms; 1 H2O molecules
Example of Finding Mass Composition
-A 0.358 g sample of chromium reacts with oxygen to from 0.523 g of metal oxide mass % Cr= 0.358 g/0.523 g X 100%= 68.5%
Example of Converting between grams & moles
-Calculate the # of moles of carbon in 0.58 g diamond Given: 0.58 g Find: mol C g C---> mol C 0.58 g C X 1 mol/12.01g C= 4.8 X10 -2 mol C
Example of Mass % Composition from a Chemical Formula
-Calculate the mass % of Cl in C2Cl4F2, freon-114 Given: C2Cl4F2 Find: Mass % Cl chemical formula---> mass % Cl mass % Cl= 4 X 35.45 g/203.8 g X 100%= 69.58%
Example of Converting Between Grams & Moles of a Compound
-Calculate the mass in grams of 1.75 mol of water Given: 1.75 mol H2O Find: g H2O mol H2O---> g H2O 1.75 mol H2O X 18.02 g H2O/ 1 mol H2O= 31.5 g H2O
Example of Converting # of Atoms to Moles
-Convert 1.1 X 10 22 Ag atoms to moles of Ag Given: 1.1 X 10 22 Ag atoms Find: moles of Ag Ag atoms---> moles of Ag 1.1 X 10 22 Ag atoms X 1 mol Ag/6.022 X 10 23 Ag atoms= 1.8 X 10 -2 mol of Ag
Example of Converting Moles to # of Atoms
-Convert 3.5 mol He to the # of He atoms Given: 3.5 mol He Find: He atoms mol He---> He atoms 3.5 mol He X 6.022 X 10 23/1 mol He= 2.1 X10 24 He atoms
Examples of Converting Between Grams of a Compound & Grams of a Constituent Element
-Find the mass of Na in 15 g of NaCl Given: 15 g Na Cl Find: g Na g NaCl---> mol NaCl---> mol Na---> g Na 15 g NaCl X 1 mol NaCl/58.44 g NaCl X 1 mol Na/ 1 mol NaCl X 22.99 g Na/ 1 mol Na= 5.9 g Na
Examples of Converting Between Moles of a Compound & Moles of a Constituent Element
-Find the number of moles of oxygen in 1.7 mol CaCO3 Given: 1.7 mol CaCO3 Find: mol O mol CaCO3---> mol O 1.7 mol CaCO3 X 3 mol O/1 mol CaCO3= 5.1 mol O
Conversion of Fractional Subscripts to Whole Numbers
-If after dividing by the smallest number of moles, the subscripts are not whole numbers, multiply all the subscripts by a small whole number to arrive at whole-number subscripts
Example of Mass % Composition as conversion factors
-The FDA recommends that adults consume less than 2.4 g of Na per day. How many g of NaCl can you consume & still be within the FDA guidelines? *NaCl is 3.9% Na by Mass Given:2.4 g Na 39 g Na/100 g NaCl Find: g NaCl g Na---> g NaCl 2.4 g Na X 100g NaCl/39 g Na= 6.2 g NaCl
Examples of Chemistry & Health: Fluoridation of Drinking Water
-What is the mass % composition of F- in NaF? -How many g of NaF should be added to 1500 L of H2O to fluoridate it at a level of 1.0 mg F-/L
Example of Converting Between # of Molecules & Mass of a Compound
-What is the mass of 4.78 X 10 24 NO2 molecules? Given: 4.78 X 10 24 NO2 molecules Find: g NO2 NO2 molecules---> mol NO2---> g NO2 4.78 X 10 24 NO2 molecules X 1 mol NO2/ 6.022 X 10 23 NO2 molecules X 46.01 g NO2/ 1 mol NO2= 365 g NO2
Calculating an Empirical formula from reaction
-a 3.24 g sample of titanium reacts with oxygen to form 5.40 g of the metal oxide. What is the empirical formula of the metal oxide Given: 3.24 g Ti 5.40 g metal oxide Find: empirical formula *the difference is the mass of oxygen that combined with the titanium Mass oxygen= 5.40 g Ti & O- 3.24 g Ti= 2.16 g O CH2O X n
Empirical Formulas from Mass % Composition: Chemical--->Mass % Composition
-an empirical formula gives only the smallest whole-number ratio of each type of atom in a compound, not the specific number of each type of atom in a molecule a) the molecular formula is always a whole-number multiple of the empirical formula EX: the molecular formula for hydrogen peroxide is H2O2 & its empirical formula is HO -Molecular formula= empirical X n, where n= 1,2,3 n=2 for H2O2
Example of Mass % Composition from a Chemical Formula
-based on the chemical formula, the mass % of the element Cl in Compound CCl2F2: mass % Cl= 2 X molar mass Cl/ molar mass CCl2F2 X 100%
Calculating Molecular Formulas for Compounds: Fructose
-find the molecular formula for fructose (a sugar found in fruit) from its empirical formula, CH2O, & its molar mass 180.2 g/mol Empirical formula molar mass= 1(12.01) + 2 (1.01)+ 16.00 =30.03 g/mol n+ 180.2 g/mol/ 30.03 g/mol= 6 So the molecular formula fo fructose is CH2O X ^= C6H12O6
Chemistry & Health: Fluoridation of Drinking Water
-fluoride strengthens tooth enamel, which prevents tooth decay -too much fluoride can cause teeth to become brown & spotted, a condition known as dental fluorosis a)Extremely high levels can lead to skeletal fluorosis -the scientific consensus is that, like many minerals, fluoride shows health benefits at certain levels-- about 1-4 mg/day for adults-- but can have detrimental effects at higher levels -adults who drink between 1 & 2 L of water per day would receive the beneficial amounts of fluoride from the water -Fluoride is often added to H2O as NaF
Fractional Subscripts
look at picture
mass % formula
mass % of element X= mass of X in sample of compound/ mass of sample of compound X 100%
Counting Molecules by the Gram
-for elements, the molar mass is the mass of 1 mol of atoms of that element -for componds, the molar mass is the mass of 1 mol of molecules or formula units of that compounds a) Ionic compounds do not contain individual molecules -We convert between the mass of a compound & moles of the compound, & then we calculate the # of molecules (or formula units from moles)
Example of Converting between grams & # of of atoms
-how many aluminum atoms are in an aluminum can with a mass of 16.2 g? Given: 16.2 g Al Find: Al atoms g Al---> mol Al---> # of atoms 16.2 g Al X 1 mol Al/ 26.98 g Al X 6.022 X 10 23 Al atoms/ 1 mol Al= 3.62 X 10 23 Al atoms
Avagadro's number
-one mole of anything is 6.022 X 10 23 units of that thing -this # is called (blank), named after Amadeo (blank) a) one mole of marbles corresponds to 6.022 X 10 23 marbles b) one mole of sand grains corresponds to 6.022 X 10 23 sand grains
How much sodium?
-sodium is an important dietary mineral that we eat in our food, primarily as sodium chloride (table salt) -sodium is involved in the regulation of body fluids & eating too much of it can lead to high blood pressure.
Counting by Weighing: Nails by the Pound
-some hardware stores sell nails by the pound, which is easier than selling them by the nail -this problem is similar to asking how many atoms are in a given mass of an element
How much sodium in sodium chloride?
-the FDA recommends a person consume less than 2.4 g (2400 mg) of sodium per day -the mass of sodium that we eat is not the same as the mass of Sodium Chloride that we eat -How many grams of sodium chloride can we consume & still stay below the FDA recommendation for sodium?
How much sodium in sodium chloride? (continued)
-the chemical composition of sodium chloride is given in its formula, NaCl a) there is one Na to every Cl ion -since the masses of sodium and chloride are different, the relationship between the mass of sodium & the mass of sodium chloride is not clear from the chemical formula alone -we need to calculate the amount of a constituent element in a given amount of a compound
Chemical Formulas as Conversion Factors
-the formula for carbon dioxide, CO2, means there are two oxygen atoms per one CO2 molecule a) we write this as follows: 2 O atoms: 1 dozen CO2 molecules b) Similarly: 2 dozen O atoms: 1 dozen CO2 molecules c) 2 mol O: 1 mol CO2
Tip
-the information in a given chemical formula, along with Atomic & Formula Masses, can be used to calculate the amount of a constituent element in a compound
Tips (part 3)
-the lighter the atom, the less mass in 1 mol of that atom
Mass % Composition of Compounds
-the mass % composition, or mass %, of an element is the element's % of the total mass of the compound -we can use mass % composition as a conversion factor between grams of constituent element & grams of a compound a) The mass % composition of Na in NaCl is 39% or 39 g Na: 100 g NaCl or 39g Na/100 g NaCl or 100g NaCl/39 g Na -these fractions are conversion factors between g Na & g NaCl
The mass of 1 mol of atoms of an element is its Molar mass
-the mass of 1 mol of atoms changes for different elements a) 32.07 g sulfur= 1 mol sulfur= 6.022 X 10 23 S atoms b) 12.01 g carbon= 1 mol carbon= 6.022 X 10 23 C atoms c) 6.49 g lithium= 1 mol lithium= 6.022 X 10 23 Li atoms
Converting Between Grams & Moles of a Compound Requires the Molar Mass of the Compound
-the molar mass of a compound in g/mol is numerically equal to the formula mass of the compound in amu a)the formula mass for a compound is the sum of the atomic masses of all the atoms in a chemical formula
The size of the mole is measured quantity
-the numerical value of the mole is defined as being equal to the # of atoms in exactly 12 g of pure carbon-12 a)this definition of the mole establishes a relationship between mass (grams of carbon) & # of atoms (Avogadro's #) b) this relationship allows us to count atoms by weighing them
Tips (part 2)
-the weight of 1 dozen nails changes for different nails -the weight of one mole of atoms changes for different elements
Molecular Formulas for Compounds Rules
-use the molar mass (which is given) & empirical formula molar mass (which you can calculate based on the empirical formula) -to determine n (the integer by which you must multiply the empirical formula to get the molecular formula -Multiply the subscripts in the empirical formula by n to arrive the molecular formula
Calculating Empirical Formula from Experimental Data: Decomposition of Water
-we decompose a sample of H2O in the lab & find that it produces 3.0 g of H & 24 g of oxygen How many moles of each element are formed during the decomposition of H2O? element mol H 3.0g H X 1 mol H/1.01 g H= 3.0 mol H mol O 24 g O X 1 mol/16.00 g O= 1.5 mol O 3.0 mol H/1.5= 2 1.5 mol/1.5=1 Empirical Formula is H2O
Counting by Weighing: Atoms by the Gram
-with atoms, we must use their mass as a way to count them a) atoms are too small & too numerous to count individually -even if you could see atoms & counted them 24 hours a day as long as you lived, you would barely begin to count the # of atoms in something as small as a grain of sand -with nails, we used a dozen as a convenient # in our conversions a) A dozen is too small to use with atoms -we need a larger # because atoms are so small a) the chemist's "dozen" is called the mole (mol) 1 mol=6.022 X10 23
Obtaining an Empirical Formula from Experimental Data
-write down (or calculate) as given the masses of each element present in a sample of the compound. If given mass % composition, assume a 100-g sample & calculate the masses of each element from given percentages -convert each of the masses in step 1 to moles by using the appropriate molar mass for each element as a conversion factor -write down a pseudo-formula for the compound, using the moles of each element (From step 2) as subscripts -divide all subscripts in the formula by the smallest subscripts -if the subscript are not whole numbers, multiply all the subscripts a small whole number to arrive at whole number subscripts
Mole Relationships from Chemical Formula
1 mol CCl4: 4 mol Cl -The relationship inherent in a chemical formula allow us to convert between moles of the compound & the moles of a constituent element (& vice versa)
How many Leaves on 14 Clovers (3-Leaf Clover Analogy)
3 leaves: 1 clover clovers--->leaves 14 clovers X 3 leaves/1 clover= 42 leaves
Nails by the Pound (example)
A customer buys 2.60 lb of medium-sized nails, & a dozen of these nails weigh 0.150 lb. How many nails did the customer buy? lb nails--->dozen nails---> # of nails 2.60 lbs nails X 1 doz nails/0.150 lb nails X 12 nails/1 doz nails= 208 nails
Example
How much iron is in a given amount of iron ore? How much Cl is in a given amount of chlorofluorcarbon?
