Composition of Substances and Solutions Atomic Mass and Formula Mass; - PowerPoint Presentation

Composition of Substances and Solutions Atomic Mass and Formula Mass; Mole & Molar Mass; Percent Composition of Compounds; Determination of Empirical & Molecular Formulas; Molarity Other Units for Solution Concentrations

Atomic Masses Absolute masses of atoms cannot be obtained – too small to measure the mass directly; Atomic mass is expressed as relative mass – masses relative to a chosen standard or reference . Carbon-12 was chosen as reference, and assigned an atomic mass of 12 amu exactly; Masses of other atoms are relative to that of carbon-12 atom; Relative atomic masses are determined using mass spectrophotometer ;

A Schematic Diagram of Mass Spectrophotometer

Isotope Mass of CO2

Formula Mass and the Mole Concept Atomic mass : The average atomic mass of an element is the weighted average atomic mass of all stable isotopes, taking into consideration the abundance of those isotopes of that element. Formula Mass: The mass of a molecule or compound is the sum of atomic masses of all the atoms represented in the substance ’ s chemical formula.

Mass Spectrum of Chlorine:1) Indicate 2 isotopes of chlorine, with relative masses of 35 amu and 37 amu ; 2) Relative abundance: Cl-35 ~75%) & Cl-37 ~25%

Mass Spectrum of Mercury(Shows a total of 7 isotopes with relative masses and abundances)

Calculating Average Atomic Masses from the Isotope Masses and Abundances Example-1 : Copper is composed of two naturally occurring isotopes: copper-63 (with mass = 62.93 amu ; abundance = 69.09%) and copper-65 (with mass = 64.93 amu ; abundance = 30.91%;). Average atomic mass of copper: = (62.93 amu x 0.6909) + (64.93 amu x 0.3091) = 63.55 amu

Calculating Average Atomic Masses from the Isotope Masses and Abundances Exercise-#1 : Chlorine has two stable naturally occurring isotopes: chlorine-35 (with mass = 34.9689 amu ; abundance = 75.76%) and chlorine-37 (with mass = 36.9659 amu ; abundance = 24.24%). Calculate the average atomic mass of chlorine? Atomic mass of chlorine = (34.9689 amu x 0.7576) + (36.9659 amu x 0.2424) = 35.45 u (as given in the periodic table)

Calculating Average Atomic Mass Exercise-#2 : Magnesium has three stable isotopes with the following masses and natural abundances: Magnesium-24 (mass = 23.9850 amu ; abundance = 79.00 %; Magnesium-25 (mass = 24.9858 amu ; abundance = 10.00 %, and Magnesium-26 (mass = 25.9826 amu ; abundance = 11.00 %) . Calculate the average atomic mass of magnesium . (Answer: 24.31 amu )

Calculating Relative Abundance of Isotopes Exercise-#3: Boron has two stable isotopes: boron-10 (mass = 10.013 amu ) and boron-11 (mass = 11.009 amu ). If the average atomic mass of boron is 10.81 amu , calculate the relative abundance of each isotope. ( Answer: boron-10 = 20.0%; boron-11 = 80.0%)

Calculating Number of Atoms from Mass Example-2 : The atomic mass of carbon is 12.01 amu . (a) How many carbon atoms are present in a 5.00-g carbon? (b) How many carbon atoms are present in a 12.01-g carbon ? (1 amu = 1.6605 x 10 -24 g) Solution: First, we find what is 12.01 amu in gram: 12.01 amu x = 1.994 x 10-23 g Then, divide the mass of sample by the mass of a C-atom in gram: (a) Number of C-atoms = 5.00 g x = 2.51 x 1023 atoms; (b) # of C-atoms = 12.01 g x = 6.023 x 1023 atoms  

Calculating Number of Atoms from MassExercise-#4: The atomic mass of silicon is 28.09 amu . (a) How many silicon atoms are present in a 5.00-g silicon ? (b) How many silicon atoms are present in 28.09 g of silicon? (1 amu = 1.66054 x 10 –24 g)

Atomic Mass & Molar Mass Examples: Element Atomic mass Molar mass Carbon 12.01 u 12.01 g/mol Oxygen 16.00 u 16.00 g/mol Aluminum 26.98 u 26.98 g/mol Silicon 28.09 u 28.09 g/mol Gold 197.0 u 197.0 g/mol

Molecular Mass, Formula Mass & Molar Mass Molecular mass = mass of one molecule (in u ); Molar mass = mass of one mole of element or compound (in g/mol) = sum of atomic masses; Examples: Molecular Mass Molar Mass (g/mol) N 2 28.02 u 28.02 H 2 O 18.02 u 18.02 C8H18 114.22 u 114.22 C12H22H11 342.30 u 342.30

Formula Mass for Covalent Molecules For covalent molecules the molecular formula represents the number and types of atoms composing a single molecule of that substance. The formula mass is also referred to as molecular mass. Consider the compound caffeine: C 8 H 10 N 4 O 2 What is the molar mass of the compound? What is the percentage composition (by mass) of each element in caffeine?

Formula Mass for Ionic Compounds For ionic compounds, the chemical formula represents the types of cations and anions and the ratio in which they combine to achieve electrically neutral matter. The chemical formula does NOT represent the composition of a discrete molecule. Consider the compound, sodium chloride (NaCl)

The Mole Reporting the number of atoms, molecules, or ions in a sample is not practical because atoms are so small. Instead chemists use the unit called the mole to represent the quantity of substances. The mole is defined as the amount of a substance containing the same number of discrete entities (such as atoms, molecules, or ions) as the number of atoms in a sample of pure carbon-12 weighing exactly 12 g.

The Mole A sample of carbon-12 isotope that weighs exactly 12 g contains 6.022 x 10 23 atoms. Avogadro’s Number = 6.022 x 10 23 A mole is a quantity that contains the Avogadro’s Number of items (atoms, molecules, ions, etc.);

Avogadro ’ s Number and the Mole The term “ mole ” is analogous to the word dozen; Dozen – A collection of 12 items. Mole – A collection of 6.022 x 10 23 items. The relationship that 1 mole = 6.022 x 1023 items is an extremely important and useful conversion factor in chemistry.

1 Mole of anything is 6.022 x 10 23 items 1 mole of silicon = 6.022 x 10 23 Si atoms 1 mole of O 2 molecules = 6.022 x 10 23 O 2 molecules 1 mole of H 2 O = 6.022 x 10 23 H 2 O molecules 1 mole of Na + ions = 6.022 x 1023 Na+ ions 1 mole of electrons = 6.022 x 1023 electrons1 mole of pennies = 6.022 x 1023 pennies

The Mole and Avogadro ’ s Number allows us to count atoms and molecules by mass The atomic mass of an element in the periodic table: is equivalent to the average mass of one atom of that element in atomic mass units (amu). but also equivalent to the mass of one mole of that element in grams. In the lab we rarely work in amu because we deal with large collections of atoms and molecules. It is much more practical to work in grams .

One mole of any element contains 6.022 x 10 23 atoms of that element. T he mass of one mole of different elements are not the same because the masses of the individual atoms are different. Each of the above samples contains 1.00 mole of atoms (or 6.02 × 10 23 atoms) of the element . (credit: modification of work by Mark Ott)

Just how big is 6.022 x 10 23 ? In order to obtain a mole of sand grains (6.022 x 10 23 grains of sand), it would be necessary to dig the entire surface of the Sahara desert (which has an area slightly less than that of the United States) to a depth of 6 feet. If you had a mole of dollars (6.022 x 10 23 dollars), and if you spend this money at the rate of one billion (1 x 10 9 ) dollars per second, it would take you over 19 million years to spend all of the money.

The number of molecules in a single droplet of water is roughly 200 billion times greater than the number of people on earth. (credit: “ tanakawho ” /Wikimedia commons)

Molar Mass ( MM ) The molar mass ( MM ) , of an element or a compound is the mass in grams of 1 mole of that substance . Examples: Molar Mass (g/ mol ) Al 26.98 H 2 O 18.02 C 12 H22H11 342.30

Calculating Molar Mass Example-3 : Consider the compound caffeine, C 8 H 10 N 4 O 2 . What is the molar mass of the compound? = ( 8 x 12.01) + (10 x 1.008) + ( 4 x 14.01) + (2 x 16.00) = 194.2 g/ mol ; What is molar mass of ammonium phosphate, (NH4)3PO4? = (3 x 14.01) + (12 x 1.008) + (1 x 30.97) + (4 x 16.00) = 149.1 g/mol

Measuring Amounts The simplest way to measure the amount of an element or a compound is by weight. In chemistry we typically express the amount of a substance in grams or moles.

Calculation of Moles from Mass The quantity in moles of a substance can be calculated from the sample mass in grams and the molar mass of the substance as follows: Mole =  

Calculating Moles from Sample Mass Example-4 : How many moles of caffeine, C 8 H 10 N 4 O 2 , are present in 5.00 g of this compound? Solution: Molar mass of caffeine = (8x12.01)+(10x1.008)+(4x14.01)+(2x16.00) = 194.2 g/ mol ; Mole of caffeine in a 5.00-g sample = 5.00 g x = 0.0257 mol; 

Calculating mole of Ammonium Phosphate from sample mass Example-5: How many moles of ammonium phosphate, (NH 4 ) 3 PO 4 , are present in 5.00 x 10 2 g of this compound? Solution: Molar mass of ammonium phosphate = (3x14.01) + (12x1.008) + (1x30.97) + (4x16.00) = 149.1 g/ mol ; Moles of (NH 4 ) 3 PO 4 in a 5.00 x 102-g sample = 5.00 x 102 g x = 3.35 moles 

Calculating Moles and Number of Atoms Exercise-#5 : Calculate the molar mass of sucrose, C 12 H 22 O 11 . How many moles of sucrose are present in 5.00 g of sucrose? How many sucrose molecules are present in 5.00 g of sucrose? How many atoms of carbon, hydrogen, and oxygen, respectively, are present in a 5.00-g sample? ( Answer : (a) 342.3 g/ mol ; (b) 0.0146 mol ; (c) 8.80 x 10 21 molecules; (d) 1.06 x 10 23 C-atoms; 1.94 x 1023 H-atoms; 9.68 x 1022 O-atoms)

Calculating Moles and Number of Atoms Exercise-#6 : Calculate the molar mass of aluminum sulfate, Al 2 (SO 4 ) 3 . How many moles of Al 2 (SO 4 ) 3 are present in 25.0 g of this compound? How many Al 2 (SO 4 ) 3 units are present in 25.0 g of this compound?How many aluminum, sulfur, and oxygen atoms, respectively, are present in a 25.0-g sample? (Answer: (a) 342.14 g/mol; (b) 0.0731 mol; (c) 4.40 x 1022 units; (d) 8.80 x 1022 Al-atoms; 1.32 x 1023 S-atoms; 5.28 x 1023 O-atoms)

Calculating numbers of atoms or molecules from Mass The number of atoms or molecules in a sample may be calculated from the sample mass in grams, the molar mass of the substance, and the Avogadro’s number. This involves converting the mass into mole and then multiplying the moles with Avogadro’s number . Number of atoms or molecules = x  

Calculating Number of Atoms or Molecules Example-5 : How many silicon atoms are present in a 5.00-g sample of pure silicon? Molar mass of silicon = 28.09 g/ mol ; Number of Si-atoms in 5.00 g of silicon = 5.00 g Si x x = 1.07 x 10 23 Si atoms  

Calculating Number of Atoms or Molecules Example-6 : How many H 2 O molecules are present in a 1.00-g sample (~1 mL) of water? Solution: Molar mass of water = 18.02 g/ mol ; Number of H 2 O molecules in 1.00 g of water: = 1.00 g x x = 3.34 x 1023 molecules 

Percent Composition of a Compound What is the percent composition by mass of caffeine, C 8 H 10 N 4 O 2 ? (molar mass = 194.2 g/ mol ) Mass percent of carbon = x 100% = 49.47%Mass percent of hydrogen = x 100% = 5.19%Mass percent of nitrogen = x 100% = 28.86 % Mass percent of oxygen = x 100% = 16.48% Total percent (by mass) = 100%  

Calculation of Percent Composition What is the percent composition by mass of caffeine, C 8 H 10 N 4 O 2 ? (molar mass = 194.2 g/ mol ) Mass percent of Carbon = (96.08 g / 194.2 g) x 100% = 49.47%Mass percent of Hydrogen = (10.08 g/194.2 g) x 100% = 5.19%Mass percent of Nitrogen = (56.04 g/194.2 g) x 100% = 28.86%Mass percent of Oxygen = (32.00 g/194.2 g) x 100% = 16.48%Total percent (by mass) = 100%

Percent Composition of a Compound What is the percent composition by mass of ammonium phosphate, (NH 4 ) 3 PO 4 ? (molar mass = 149.1 g/ mol ) Mass percent of nitrogen = x 100% = 28.19% Mass percent of hydrogen = x 100% = 8.12%Mass percent of phosphorus = x 100% = 20.77% Mass percent of o xygen = x 100% = 42.92% Total percent (by mass) = 100%  

Calculation of Percent Composition What is the percent composition by mass of ammonium phosphate, ( NH 4 ) 3 PO 4 ? (molar mass = 149.1 g/ mol ) Mass percent of N = (42.03 g / 149.1 g) x 100% = 28.19% Mass percent of H = (12.10 g / 149.1 g) x 100% = 8.12%Mass percent of P = (30.97 g/149.1 g) x 100% = 20.77%Mass percent of O = (64.00 g/149.1 g) x 100% = 42.92%Total percent (by mass) = 100%

Determination of Empirical and Molecular Formulas Empirical Formula A formula that shows the simples whole number ratio of moles of atoms . Examples : MgO, Cu 2 S, CH 2 O, Al 2 C 3 O 9 . etc. Molecular Formula A formula that shows the actual number of atoms of each type of element in a molecule. Examples : C 4H10, C6H6, C6H12O6, N2H4, P4O10, etc.

Determination of Empirical Formula Example-7: A 5.00 sample of an oxide of phosphorus is found to contain 2.18 g of phosphorus and 2.82 g of oxygen. Calculate the empirical formula of the compound. Solution: Mol of P = 2.18 g x = 0.0704 mol ; Mol of O = 2.82 g x = 0.176 mol ; Mole ratio : = = 2.5/1 = 5/2 Empirical formula = P2O5  

Determination of Empirical Formula Example-8 : A compound made up of carbon, hydrogen, and oxygen has the following composition (by mass percent): 68.1% C, 13.7% H, and 18.2% O. Determine its empirical formula. Solution : Treat mass percent like mass and calculate mole of each element in 100-g sample of the compound. Mole of C = 68.1 g x (1 mol C/12.01 g) = 5.67 mol ; Mole of H = 13.7 g x (1 mol H/1.008 g) = 13.6 mol ; Mole of O = 18.2 g x (1 mol O/16.00 g) = 1.14 mol ; Divide throughout by the mole of O (the smallest mole value): 5.67 mol C/1.14 = 4.97 ~ 5; 13.6 mol H/1.14 = 11.9 ~ 12; 1.14 mol O/1.14 = 1; Empirical formula = C5H12O

Determination of Empirical Formula Example-9 : A 2.32-g sample of a compound composed of carbon, hydrogen, and oxygen is completely combusted to yield 5.28 g of CO 2 gas and 2.16 g of water. (a) Calculate the composition of the compound (in mass percent ); (b) Determine its empirical formula. Solution : Determine the mass of C, H, and O, respectively, in the sample: Mass of C = 5.28 g CO 2 x (12.01 g C/44.01 g CO 2 ) = 1.44 g; Mass of H = 2.16 g H 2 O x (2 x 1.008 g/18.02 g H 2 O) = 0.242 g; Mass of O = 2.32 g sample – 1.44 g C – 0.24 g H = 0.64 g; Calculate the mass percent of each element: Mass % of C = (1.44 g C/2.32 g sample) x 100% = 62.1% ; Mass % of H = (0.242 g H/2.32 g sample) x 100% = 10.4% ; Mass % of O = 100 – 62.1% C – 10.4% H = 27.5% Composition (by mass %): 62.1% C; 10.4% H, and 27.5% O;(continue next slide for empirical formula determination)

Determination of Empirical Formula Example-9 (continued): Calculate mole of each element from mass percent: Mole of C = 62.1 g C x (1 mol /12.01 g) = 5.17 mol Mole of H = 10.4 g H x (1 mol /1.008 g) = 10.3 mol Mole of O = 27.5 g O x (1 mol /16.00 g) = 1.72 mol Dividing throughout by smallest mole of O yield a simple ratio: 5.17 mol C/1.72 = 3 mol C; 10.3 mol H/1.72 = 6 mol H; 1.72 mol O/1.72 = 1 mol O;Simple mole ratio of the elements: 3 mol C : 6 mol H : 1 mol OEmpirical formula: C3H6O

Molecular Formula Molecular formula of a compound is determined from the empirical formula and the molar mass of the compound; the latter is determined independently. If empirical formula = C x H y O z Molecular formula = ( C x H y O z ) n Where n = ; (n is an integer) 

Determination of Molecular Formula Example-10 : A compound with empirical formula C 3 H 6 O has molecular mass of 116 amu . What is the molecular formula? Solution : Empirical formula mass = (3 x 12.01) + (6 x 1.008) + 16.00 amu = 58.1 amu n = = 2; Molecular formula = C6H12O2 

Determination of empirical and molecular formula of phosphorus oxide Exercise-#7 : A 2.50-gram sample of phosphorus is completely burned in air, which yields 5.72 g of product composed of only phosphorus and oxygen. In a separate analysis, the compound was found to have molar mass of 284 g/mol. (a) Determine the empirical formula and molecular formula of the compound. (b) What is the name of the compound? (c) Write an equation for the combustion of phosphorus . Answer : (a) Empirical formula = P 2 O 5 ; molecular formula = P 4 O 10 ; (b) Tetraphosphorus decoxide; (c) 4P(s) + 5 O2(g)  P4O10(s)

Solution Concentrations The concentration of a solution may be expressed in: Molarity, mass percent , volume percent , mass-volume percent , ppm & ppb .

Solution Concentrations Mass-volume percent = x 100% Parts per million (ppm) = x 10 6 Parts per billion (ppb) = x 10 9  

Molar Concentration Example-11 : 9.00 g of sodium chloride, NaCl , is dissolved in enough water to make 250.0 mL of salt solution. Calculate the molarity of NaCl . Mole of NaCl = 9.00 g NaCl x = 0.154 mol ; Molarity of NaCl = = 0.616 M  

Calculation of Solute Mass in Solution Example-12 : How many grams of NaOH are required to prepare 750.0 mL of 2 .0 M NaOH solution? ( Molar mass of NaOH = 40.0 g/ mol ) ? Answer : Mole of NaOH needed = 0.7500 L x = 1.5 mol;Mass of NaOH needed = 1 5 mol x = 60. g 

Preparing of Solutions from Pure Solids From the volume (in liters) and the molarity of solution, calculate the mole and mass of solute needed; Weigh the mass of pure solute accurately; Transfer solute into a volumetric flask of appropriate size; Add deionized water to the volumetric flask, well below the narrow neck, and shake well (or use a magnetic stirrer) to dissolve the solute; When solid is completely dissolved, add more deionized water to fill the flask to the mark and mix the solution well. ( Note : If the solution is very warm due to exothermic reaction, let it cools down to room temperature before adding more water to the “mark”.)

Preparing Solution from Solid Example-13 : Explain how you would prepare a 500.0 mL of 0.154 M NaCl solution. Calculate the mole and mass of NaCl needed: Mole of NaCl = 0.5000 L x (0.154 mol /L) = 0.0770 mol Mass of NaCl = 0.0770 mol x (58.44 g/ mol) = 4.50 gPreparing the solution: Weigh accurately 4.50 g of NaCl and transfer solid into 500-mL volumetric flask. Fill the flask half way with distilled water, shake well until all solid has dissolved. Add deionized water to fill the flask to the 500-mL level, place a stopper on the flask and mix the solution well by inverting and shaking the flask several times.

Calculating MolarityExercise-#8 : 10.55 g of silver nitrate, AgNO 3 ( molar mass = 169.9 g/ mol ), is dissolved in enough water to make a 500.0-mL solution. What is the molar concentration of AgNO 3 ? (Answer: 0.1242 M )

Calculating Mass of Solute Exercise-#9 : How many grams of glucose, C 6 H 12 O 6 , are needed to prepare 473.2 mL of 0.3065 M glucose solution? Briefly explain the method you would use to prepare this solution. ( molar mass of glucose = 180.16 g/ mol ) (Answer: 26.13 g)

Preparing Solution from Stock Calculate volume of stock solution needed using the formula: M i V i = M f V f ( i = initial; f = final) Measure accurately the needed volume of stock solution and transfer to a volumetric flask of appropriate size; Dilute stock solution with distilled water to the required volume (or to the “mark” on volumetric flask) Mix solution well. ( Note : for diluting concentrated acid, place some deionized water in the flask (to about a quarter full), add the required volume of concentrated acid, and then add more deionized water to the required volume.)

Preparing Solution from Stock Example-14 : Explain how you would prepare 1.0 L of 3.0 M H 2 SO 4 solution from concentrated H 2 SO 4 , which is 18 M . Calculate volume of concentrated H 2 SO 4 needed: Vol. of conc. H 2 SO4 = (1.0 L x 3.0 M/18 M) = 0.17 L = 170 mLPreparing the solution: Place about 200 mL of deionized water in the 1-liter volumetric flask. Measure accurately 170 mL of concentrated H2SO4 and transfer carefully to the volumetric flask containing some deionized water. Allow solution to cool down to room temperature; then fill the flask to the 1-liter mark with more distilled water; Mix solution well by placing a stopper on the flask and inverting the flask back and forth several times.

Calculating Volume of Stock Solution Exercise-#10 : 35.0 mL of 6.0 M NaOH solution is added into a 500.0-mL volumetric flask, and enough deionized water is then added to make the final solution to 500.0 mL. Calculate the molar concentration of NaOH in the dilute solution. (Answer: 0.42 M )

Calculating Volume of Stock Solution Exercise-#11 : How many milliliters (mL) of concentrated sulfuric acid, H 2 SO 4 , are needed to prepare 5.0 x 10 2 mL of 3.0 M H 2 SO 4 solution? ( Molarity of concentrated H 2 SO 4 is 18 M )Explain the safe method you would use to prepare this solution. (Answer: (a) 83 mL)

Other Units for Solution Concentration  

Calculating of Mass Percent Example-15 : A sugar solution contains 25.0 g sugar dissolved in 100.0 g of water. What is the mass percent of sugar in the solution? Answer: Percent sugar = x 100% = 20.0% (by mass)  

Calculating of Mass Percent Example-16: Seawater contains 3.5% NaCl , by mass. How many grams of salt are present in 5.00 gallons of seawater ( sw )? Solution: (1 gallon = 3785 mL; assume density = 1.00 g/mL) Mass of seawater = 5.00 gall x x = 1.89 x 10 4 g Mass of salt = 1.89 x 10 4 g sw x = 660 g NaCl 

Calculation of Mass Percent Exercise-#12 : 20.00 g of sugar and 5.00 g of salt ( NaCl ) are dissolved in 100.0 g of water. Calculate the mass percent of sugar and salt, respectively, in the solution. (Answer: 16.00% sugar; 4.00% salt)

Calculating Volume Percent Example-17 : 150 mL of methanol, 120 mL of acetone and 230 mL of water are mixed to form a homogeneous 5.0 x 10 2 mL solution. Calculate the percentage (by volume) of each component in the mixture. Solution : Percent methanol = x 100 = 30.% Percent acetone = x 100 = 24% Percent water = x 100 = 46%  

Calculation Mass Percent Exercise-#13 : Gasohol is a mixture containing 10.% ethanol (C 2 H 5 OH) and 90.% gasoline (by volume). The density of ethanol and gasoline are 0.79 g/mL and 0.70 g/mL, respectively. What are the composition of gasohol by mass percent? (Answer: 11% ethanol; 89% gasoline)

Calculation using Mass Percent Exercise-#14 : Gasohol contains 10.% ethanol, C 2 H 5 OH, and 90.% gasoline (by volume). The density of ethanol is 0.79 g/ mL. How many grams of ethanol are present in 1.00 gallon of gasohol? (1 gallon = 3.785 L) What is the molarity of ethanol in gasohol? (Answer: (a) 3.0 x 10 2 g; (b) 1.7 M )

Calculating Mass-Volume Percent Example-18: 9.0 g of NaCl is dissolved in 473 mL of solution. What is the percent (w/v) of NaCl in the solution? 28 g of glucose is dissolved in 475 mL of solution. What is the percent (w/v) of glucose in the solution? Solutions : Percent of NaCl = x 100 = 1.9% (w/v) Percent of glucose = x 100 = 5.9% (w/v)  

ppm and ppb