Chemical Quantities Moles, % Compositions,

3 ways to measure matter Number present By Mass/Weight By Volume Count how many are there Example: number of songs on your smartphone By Mass/Weight How many kg/lb present? Example: buying produce at a supermarket By Volume How many gallons/liters present? Example: buying gas

Using Apples as an Example Remember Conversion Factors A ratio equal to 1 1 dozen apples 12 apples 2.0 kg 0.2 bushel of apples Fruit stand Often by number present 1 dozen= 12 apples Supermarket Often by mass/weight 1 dozen= 2.0 kg Orchard Often by volume 1 dozen= 0.2 bushels

What is the mass of 90. average sized apples if 1 dozen of the apples has a mass of 2.0kg? Identify what you want to change Number of apples  dozens  mass Identify your equalities (conversion factors) 1 dozen = 12 apples and 2.0 kg = 1 dozen Solve the problem using dimensional analysis 90. apples 1 dozen 2.0 kg 15 kg 12 apples 1 dozen

What is a mole? A way to determine how many particles are present in chemistry because they are way too small to count 1 mol = 6.022 x 1023 representative particles Avogadro’s Number- Italian scientist Amedeo Avogadro who helped clarify the difference between atoms and molecules Species present in a substance- this may vary depending on the substance (examples on next slide)

Examples of Representative Particles Substance Representative Particle Chemical Formula Representative Particles in 1.00 mol Copper atom Cu 6.022 X 1023 Water molecule H2O Calcium Ion ion Ca2+ Calcium Fluoride formula unit CaF2

Conversion factor possibilities for a mol 1 mol = 6.022 x 1023 rp rp= representative particle

Moles to Particles or Particles to Moles (1 Step Problem) Number of RP (atoms, molecules, f.u. Avogadro’s number

Magnesium is a light metal used in the manufacture of aircraft, automobile wheels, and tools. How many moles of Mg is 1.25 x 1023 atoms of Mg? 1.25 x 1023 atoms Mg

Atoms are your representative particles in this case Magnesium is a light metal used in the manufacture of aircraft, automobile wheels, and tools. How many moles of Mg is 1.25 x 1023 atoms of Mg? 1 mol Mg 1.25 x 1023 atoms Mg 6.022 x 1023 atoms Mg Atoms are your representative particles in this case

Magnesium is a light metal used in the manufacture of aircraft, automobile wheels, and tools. How many moles of Mg is 1.25 x 1023 atoms of Mg? 1 mol Mg 1.25 x 1023 atoms Mg 0.208 mol Mg 6.022 x 1023 atoms Mg

How many moles of Si are in 2.8 x 1024 atoms of silicon? Answer 4.65 mol Si

Propane gas (C3H8) is often used for cooking and heating Propane gas (C3H8) is often used for cooking and heating. How many carbon atoms are in 1 molecule of propane? How many hydrogen atoms are in 1 molecule of propane? Answer: Carbon- 3.83 x 1024 Answer: Hydrogen- 1.02 x 1025

Propane gas (C3H8) is often used for cooking and heating Propane gas (C3H8) is often used for cooking and heating. How many moles of carbon are in 1 mole of propane? How many moles of hydrogen are in 1 mole of propane? Answer: Carbon- 3.83 x 1024 Answer: Hydrogen- 1.02 x 1025

Molar Mass Atomic mass = mass of a single atom expressed in amu (atomic mass unit); relative values based on the isotopes; (decimals) on the periodic table Molar mass is the mass of 1 mole of an element/compound/molecule; equal to the atomic mass unit for molar mass is: g/mol (grams per mole) Example: Carbon’s atomic mass is 12.01 amu and its molar mass is 12.01 g/mol

1 MOLE

How to find the mass of a mole of a compound You must know the chemical formula Then add the atomic masses of the atoms that make up that compound Example: Sulfur Trioxide formula- SO3 1 sulfur atom and 3 oxygen atoms Sulfur molar mass = 32.06 g/mol Oxygen molar mass = 16.00 g/mol 32.06 + 16.00 + 16.00 + 16.00 = 80.06 g/mol Substituted g/mol for amu

The decomposition of hydrogen peroxide (H2O2) provides sufficient energy to launch a rocket. What is the molar mass of hydrogen peroxide? Formula = H2O2 2 hydrogens and 2 oxygens

The decomposition of hydrogen peroxide (H2O2) provides sufficient energy to launch a rocket. What is the molar mass of hydrogen peroxide? Formula = H2O2 2 hydrogens and 2 oxygens Hydrogen = 1.01 g/mol Oxygen = 16.00 g/mol

The decomposition of hydrogen peroxide (H2O2) provides sufficient energy to launch a rocket. What is the molar mass of hydrogen peroxide? Formula = H2O2 2 hydrogens and 2 oxygens Hydrogen = 1.01 g/mol Oxygen = 16.00 g/mol 1.01 + 1.01 + 16.00 + 16.00 = 34.02 g/mol

Find the molar mass of calcium nitrate. Answer- 164 g/mol

Mole to Mass Relationship Molar mass of a substance is the mass in grams per mole of that substance Use the molar mass of an element or compound to convert between the mass and the moles The units for molar mass are already set up as a conversion factor Conversion factors: g 1 mol 1 mol g Molar mass = OR

Moles to Mass or Mass to Moles (1 Step Problem) Molar mass Mass of substance of substance

Items made out of aluminum, such as aircraft parts and cookware, are resistant to corrosion because the aluminum reacts with oxygen in the air to form a coating of aluminum oxide (Al2O3). This tough, resistant coating prevents any further corrosion. What is the mass, in grams, of 9.45 moles of aluminum oxide?

Items made out of aluminum, such as aircraft parts and cookware, are resistant to corrosion because the aluminum reacts with oxygen in the air to form a coating of aluminum oxide (Al2O3). This tough, resistant coating prevents any further corrosion. What is the mass, in grams, of 9.45 moles of aluminum oxide? Solve for molar mass of Al2O3 : 2 Al = 26.98 g/mol + 26.98 g/mol = 53.96 g/mol 3 O = 16.00 g/mol + 16.00 g/mol + 16.00 g/mol = 48.00 g/mol Total= 53.96 g/mol + 48.00 g/mol = 101.96 g/mol

Items made out of aluminum, such as aircraft parts and cookware, are resistant to corrosion because the aluminum reacts with oxygen in the air to form a coating of aluminum oxide (Al2O3). This tough, resistant coating prevents any further corrosion. What is the mass, in grams, of 9.45 moles of aluminum oxide? Solve for molar mass of Al2O3 : 2 Al = 26.98 g/mol + 26.98 g/mol = 53.96 g/mol 3 O = 16.00 g/mol + 16.00 g/mol + 16.00 g/mol = 48.00 g/mol Total= 53.96 g/mol + 48.00 g/mol = 101.96 g/mol Use dimensional analysis: 9.45 mol

Items made out of aluminum, such as aircraft parts and cookware, are resistant to corrosion because the aluminum reacts with oxygen in the air to form a coating of aluminum oxide (Al2O3). This tough, resistant coating prevents any further corrosion. What is the mass, in grams, of 9.45 moles of aluminum oxide? Solve for molar mass of Al2O3 : 2 Al = 26.98 g/mol + 26.98 g/mol = 53.96 g/mol 3 O = 16.00 g/mol + 16.00 g/mol + 16.00 g/mol = 48.00 g/mol Total= 53.96 g/mol + 48.00 g/mol = 101.96 g/mol Use dimensional analysis: 101.96 g 9.45 mol 1 mol

Items made out of aluminum, such as aircraft parts and cookware, are resistant to corrosion because the aluminum reacts with oxygen in the air to form a coating of aluminum oxide (Al2O3). This tough, resistant coating prevents any further corrosion. What is the mass, in grams, of 9.45 moles of aluminum oxide? Solve for molar mass of Al2O3 : 2 Al = 26.98 g/mol + 26.98 g/mol = 53.96 g/mol 3 O = 16.00 g/mol + 16.00 g/mol + 16.00 g/mol = 48.00 g/mol Total= 53.96 g/mol + 48.00 g/mol = 101.96 g/mol Use dimensional analysis: 101.96 g 963.53 g = 964 g 9.45 mol 1 mol

Calculate the mass, in grams, of 2.50 mol of Iron (II) hydroxide. Answer- 225 g Formula then molar mass then dimensional analysis

When iron is exposed to the air, it corrodes to form red-brown rust When iron is exposed to the air, it corrodes to form red-brown rust. Rust is Iron (III) oxide. How many moles of Iron (III) oxide are contained in 92.2 g of pure Iron (III) oxide?

Figure out formula for Iron (III) oxide When iron is exposed to the air, it corrodes to form red-brown rust. Rust is Iron (III) oxide. How many moles of Iron (III) oxide are contained in 92.2 g of pure Iron (III) oxide? Figure out formula for Iron (III) oxide Fe3+ O2- Fe2O3

When iron is exposed to the air, it corrodes to form red-brown rust When iron is exposed to the air, it corrodes to form red-brown rust. Rust is Iron (III) oxide. How many moles of Iron (III) oxide are contained in 92.2 g of pure Iron (III) oxide? Figure out formula for Iron (III) oxide Fe3+ O2- Fe2O3 Calculate molar mass 2(55.85 g/mol) + 3(16.00 g/mol) = 159.70 g/mol

When iron is exposed to the air, it corrodes to form red-brown rust When iron is exposed to the air, it corrodes to form red-brown rust. Rust is Iron (III) oxide. How many moles of Iron (III) oxide are contained in 92.2 g of pure Iron (III) oxide? Figure out formula for Iron (III) oxide Fe3+ O2- Fe2O3 Calculate molar mass 2(55.85 g/mol) + 3(16.00 g/mol) = 159.70 g/mol Use dimensional analysis 92.2 g

When iron is exposed to the air, it corrodes to form red-brown rust When iron is exposed to the air, it corrodes to form red-brown rust. Rust is Iron (III) oxide. How many moles of Iron (III) oxide are contained in 92.2 g of pure Iron (III) oxide? Figure out formula for Iron (III) oxide Fe3+ O2- Fe2O3 Calculate molar mass 2(55.85 g/mol) + 3(16.00 g/mol) = 159.70 g/mol Use dimensional analysis 92.2 g 1 mol 159.70 g

When iron is exposed to the air, it corrodes to form red-brown rust When iron is exposed to the air, it corrodes to form red-brown rust. Rust is Iron (III) oxide. How many moles of Iron (III) oxide are contained in 92.2 g of pure Iron (III) oxide? Figure out formula for Iron (III) oxide Fe3+ O2- Fe2O3 Calculate molar mass 2(55.85 g/mol) + 3(16.00 g/mol) = 159.70 g/mol Use dimensional analysis 92.2 g 1 mol 0.577 mol Fe2O3 159.70 g

Calculate the number of moles in 75.0g of dinitrogen trioxide. Answer- 0.987 mol N2O3 Formula to molar mass to mol

Number of RP (atoms, molecules, f.u. Mass to Particles or Particles to Mass Mass to atoms or molecules A 2-step problem (2 steps) Number of RP (atoms, molecules, f.u. Molar mass Moles of Substance Avogadro’s Mass of substance number of substance **Hint – If you do not see moles in the problem, it is a 2 step problem** (2 steps)

Summary of Mole Conversions Use Avogadro’s Number Use Periodic Table MOLE Representative Particle Mass Calculate Molar Mass g/mol 6.022 x 1023 In grams Atom, Molecule, Formula Unit A B C IF YOU DON’T SEE “MOLES” IN THE PROBLEM- IT IS A 2 STEP PROBLEM!

Calculate the number of formula units in 78.4g of sodium chloride. This is a two step problem. First go from grams to moles, then moles to formula units.

Calculate the number of formula units in 78.4g of sodium chloride. This is a two step problem. First go from grams to moles, then moles to formula units. Sodium Chloride’s chemical formula is NaCl. Its molar mass is: 22.99 g/mol + 35.45 g/mol= 58.44 g/mol

Calculate the number of formula units in 78.4g of sodium chloride. This is a two step problem. First go from grams to moles, then moles to formula units. Sodium Chloride’s chemical formula is NaCl. Its molar mass is: 22.99 g/mol + 35.45 g/mol= 58.44 g/mol 1 mol NaCl 78.4 g NaCl 58.44 g NaCl

Calculate the number of formula units in 78.4g of sodium chloride. This is a two step problem. First go from grams to moles, then moles to formula units. Sodium Chloride’s chemical formula is NaCl. Its molar mass is: 22.99 g/mol + 35.45 g/mol= 58.44 g/mol 1 mol NaCl 78.4 g NaCl 6.022 x 1023 fu 58.44 g NaCl 1 mol NaCl

Calculate the number of formula units in 78.4g of sodium chloride. This is a two step problem. First go from grams to moles, then moles to formula units. Sodium Chloride’s chemical formula is NaCl. Its molar mass is: 22.99 g/mol + 35.45 g/mol= 58.44 g/mol 1 mol NaCl 78.4 g NaCl 6.022 x 1023 fu 8.34 x 1023 fu 58.44 g NaCl 1 mol NaCl

How many grams are in 94.8 molecules of SO2? Answer: 64.06 is SO2 molar mass

Mole to Volume Relationship Avogadro’s Law: equal volumes of gases at the same temperature and pressure contain equal numbers of particles Volume of a gas can change with a change in temperature and pressure so it is usually measured at STP (standard temperature and pressure) 0oC (273 K) and 1 atm. At STP: 1 mol= 6.022 x 1023 particles = 22.4 L 22.4L is called the molar volume of a gas Molar volume is used to convert between number of moles of a gas and the volume of the gas at STP

Conversion Factors for mol and volume ratio 1 mol = 22.4 L

Sulfur dioxide (SO2) is a gas produced by burning coal Sulfur dioxide (SO2) is a gas produced by burning coal. It is an air pollutant and one of the causes of acid rain. Determine the volume, in liters, of 0.60 mol of SO2 gas at STP. Known from problem 0.60 mol

Conversion factor from gas at STP Sulfur dioxide (SO2) is a gas produced by burning coal. It is an air pollutant and one of the causes of acid rain. Determine the volume, in liters, of 0.60 mol of SO2 gas at STP. 0.60 mol 22.4 L 1 mol Conversion factor from gas at STP

Sulfur dioxide (SO2) is a gas produced by burning coal Sulfur dioxide (SO2) is a gas produced by burning coal. It is an air pollutant and one of the causes of acid rain. Determine the volume, in liters, of 0.60 mol of SO2 gas at STP. 0.60 mol 22.4 L 13 L SO2 1 mol

What is the volume of 0.960 mol CH4 at STP? At STP, how many moles are in 1.00 x 103 L of C2H6? Answers- 21.5 L CH4 - 44.6 mol C2H6

The density of a gaseous compound containing carbon and oxygen is found to be 1.964 g/L at STP. What is the molar mass?

The density of a gaseous compound containing carbon and oxygen is found to be 1.964 g/L at STP. What is the molar mass? I know: Density = 1.964 g/L 1 mol of gas at STP = 22.4L

The density of a gaseous compound containing carbon and oxygen is found to be 1.964 g/L at STP. What is the molar mass? I know: Density = 1.964 g/L 1 mol of gas at STP = 22.4L 1.964 g 1 L

The density of a gaseous compound containing carbon and oxygen is found to be 1.964 g/L at STP. What is the molar mass? I know: Density = 1.964 g/L 1 mol of gas at STP = 22.4L 22.4 L 1.964 g 1 L 1 mol

The density of a gaseous compound containing carbon and oxygen is found to be 1.964 g/L at STP. What is the molar mass? I know: Density = 1.964 g/L 1 mol of gas at STP = 22.4L 22.4 L 1.964 g 43.99 g/mol 1 L 1 mol

What is the density of krypton gas at STP? Answer- 3.74 g/L Look up Krypton’s molar mass (83.8 g/mol) and use 22.4 L

Summary of mole conversions L, mL, etc…. Summary of mole conversions atoms, molecules, formula units g, kg, etc..

Percent Composition % by mass of element = mass of element x 100 Percent composition is the percent by mass of each element in a compound Using the mass of the element you solve using the following formula % by mass of element = mass of element x 100 mass of compound Where you may have seen something like this before?

When a 13.60 g sample of a compound containing only magnesium and oxygen is broken down, 5.40g of oxygen and 8.20 g of magnesium are obtained. What is the percent composition of this compound?

When a 13.60g sample of a compound containing only magnesium and oxygen is decomposed, 5.40g of oxygen is obtained. What is the percent composition of this compound? Percent composition of oxygen: 5.40g x 100 = 39.7% 13.60g

When a 13.60g sample of a compound containing only magnesium and oxygen is decomposed, 5.40g of oxygen is obtained. What is the percent composition of this compound? Percent composition of oxygen: 5.40g x 100 = 39.7% 13.60g Percent composition of magnesium: 8.20g x 100 = 60.3% 13.60g

When a 13.60g sample of a compound containing only magnesium and oxygen is decomposed, 5.40g of oxygen is obtained. What is the percent composition of this compound? Percent composition of oxygen: 5.40g x 100 = 39.7% 13.60g Percent composition of magnesium: 8.20g x 100 = 60.3% 13.60g The compound is 39.7% oxygen and 60.3% magnesium.

Percent Composition from a Chemical Formula Percent by mass of an element Molar mass of element = x 100 Molar mass of compound

Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane.

Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane. Calculate molar mass: Carbon= 3 x 12.01 g/mol = 36.03 g/mol Hydrogen= 8 x 1.01 g/mol = 8.08 g/mol Total molar mass of propane = 36.03 + 8.08 = 44.11 g/mol

Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane. Calculate molar mass: Carbon= 3 x 12.01 g/mol = 36.03 g/mol Hydrogen= 8 x 1.01 g/mol = 8.08 g/mol Total molar mass of propane = 36.03 + 8.08 = 44.11 g/mol Percent Carbon (36.03 ÷ 44.11) x 100 = 81.8% Carbon

Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane. Calculate molar mass: Carbon= 3 x 12.01 g/mol = 36.03 g/mol Hydrogen= 8 x 1.01 g/mol = 8.08 g/mol Total molar mass of propane = 36.03 + 8.08 = 44.11 g/mol Percent Carbon (36.03 ÷ 44.11) x 100 = 81.68% Carbon Percent Hydrogen (8.08 ÷ 44.11) x 100 = 18.3% Hydrogen

Example: Hydrogen peroxide Empirical Formulas Empirical formulas are the lowest whole number ratio of the atoms or moles of elements in a compound May or may not be the same as the molecular formula Example: Hydrogen peroxide

Molecular Formula vs Empirical Formula Either the same as the empirical formula or is a simple whole number multiple of the empirical formula molecular formula empirical formula H2O2 = HO C6H12O6 = N2O6 = H2O = C2H10O2 = N3O2 = P4O10 =

The percent composition of a compound can be used to calculate the empirical formula of that compound A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula?

A compound is analyzed and found to contain 25. 9% nitrogen and 74 A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? What we know: Nitrogen = 25.9% Oxygen = 74.1%

A compound is analyzed and found to contain 25. 9% nitrogen and 74 A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? What we know: Nitrogen = 25.9% Oxygen = 74.1% Step 1: Assume 100 g and calculate moles of each, so simply change % to mass (g) for each Nitrogen = 25.9g Oxygen = 74.1g

A compound is analyzed and found to contain 25. 9% nitrogen and 74 A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? What we know: Nitrogen = 25.9% Oxygen = 74.1% Step 1: Assume 100 g and calculate moles of each, so simply change % to mass (g) for each Nitrogen = 25.9g Oxygen = 74.1g Step 2: Convert mass to moles of each Moles of Nitrogen 25.9 g 14.0 g 1 mol 1.85 mol N Moles of Oxygen 74.1 g 16.0 g 1 mol 4.63 mol O

A compound is analyzed and found to contain 25. 9% nitrogen and 74 A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? What we know: Nitrogen = 25.9% Oxygen = 74.1% Assume 100 g and calculate moles of each Step 3: Divide each molar quantity by the smallest number of moles to get 1 mole for the element with the smallest number Moles of Nitrogen 25.9 g 14.0 g 1 mol 1.85 mol N 1.85 mol = 1 mol N 1.85 mol 4.63 mol = 2.5 mol O Moles of Oxygen 74.1 g 16.0 g 1 mol 4.63 mol O

A compound is analyzed and found to contain 25. 9% nitrogen and 74 A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? What we know: Nitrogen = 25.9% Oxygen = 74.1% Assume 100 g and calculate moles of each Divide each molar quantity by the smallest number of moles to get 1 mole for the element with the smallest number Moles of Nitrogen 25.9 g 14.0 g 1 mol 1.85 mol N 1.85 mol ÷ 1.85 mol = 1 mol N 4.63 mol ÷ 1.85 mol = 2.5 mol O Moles of Oxygen Step 4: Multiply each by the lowest whole number that converts both to a whole number (if necessary) 74.1 g 16.0 g 1 mol 4.63 mol O 1 mol N x 2 = 2 mol N 2.50 mol O x 2 = 5 mol O

A compound is analyzed and found to contain 25. 9% nitrogen and 74 A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? What we know: Nitrogen = 25.9% Oxygen = 74.1% Assume 100 g and calculate moles of each Divide each molar quantity by the smallest number of moles to get 1 mole for the element with the smallest number Moles of Nitrogen 25.9 g 14.0 g 1 mol 1.85 mol N 1.85 mol ÷ 1.85 mol = 1 mol N 4.63 mol ÷ 1.85 mol = 2.50 mol O Moles of Oxygen Multiply each by the lowest whole number that converts both to a whole number 74.1 g 16.0 g 1 mol 4.63 mol O 1 mol N x 2 = 2 mol N 2.50 mol O x 2 = 5 mol O Step 5: Write empirical formula: N2O5

Calculate the empirical formula for a compound that is 67. 6% Hg, 10 Calculate the empirical formula for a compound that is 67.6% Hg, 10.8% S, and 21.6% O. Answer- HgSO4 Change % to grams Calculate moles based on molar masses Divide by lowest number of moles Multiply to get each to a whole number

Calculate the molecular formula of a compound whose molar mass is 60 Calculate the molecular formula of a compound whose molar mass is 60.0g/mol and empirical formula is CH4N.

Calculate the empirical formula mass Calculate the molecular formula of a compound whose molar mass is 60.0g/mol and empirical formula is CH4N. Calculate the empirical formula mass C= 12.0 g/mol H= 1.0g/mol x 4= 4.0 g/mol N= 14.0 g/mol Total= 30.0 g/mol

Calculate the molecular formula of a compound whose molar mass is 60 Calculate the molecular formula of a compound whose molar mass is 60.0g/mol and empirical formula is CH4N. Calculate the empirical formula mass Divide the molar mass of the compound by the empirical formula mass 60.0 g/mol 30.0 g/mol C= 12.0 g/mol H= 1.0g/mol x 4= 4.0 g/mol N= 14.0 g/mol 2 = Total= 30.0 g/mol

Calculate the molecular formula of a compound whose molar mass is 60 Calculate the molecular formula of a compound whose molar mass is 60.0g/mol and empirical formula is CH4N. Calculate the empirical formula mass Divide the molar mass of the compound by the empirical formula mass 60.0 g/mol 30.0 g/mol C= 12.0 g/mol H= 1.0g/mol x 4= 4.0 g/mol N= 14.0 g/mol 2 = Multiply the empirical formula by this value to figure out the molecular formula Total= 30.0 g/mol CH4N x 2 =

Calculate the molecular formula of a compound whose molar mass is 60 Calculate the molecular formula of a compound whose molar mass is 60.0g/mol and empirical formula is CH4N. Calculate the empirical formula mass (efm) Divide the molar mass of the compound by the efm 60.0 g/mol 30.0 g/mol C= 12.0 g/mol H= 1.0g/mol x 4= 4.0 g/mol N= 14.0 g/mol 2 = Multiply the empirical formula by this value to figure out the molecular formula Total= 30.0 g/mol CH4N x 2 = C2H8N2

Find the molecular formula of ethylene glycol, which is used as antifreeze. The molar mass is 62.0 g/mol, and the empirical formula is CH3O. Answer- C2H6O2