Mole Conversions The mole is the basic SI unit of quantity and is equal to 6.02 × 1023 of anything. Here are the useful mole conversions used in chemistry… Particles (atoms, ion, molecules, etc): 1 mole = 6.02 × 1023 particles Mass (grams, etc): 1mole = molar mass (mass of 6.02 × 1023 particles) Volume (liters, etc): 1 mole = 22.4L at STP (273 K and 101.3 kPa) The molar mass and formula mass for a compound are found in the same manner…simply add up the mass contributed by each element in the compound (found by multiplying each element’s atomic mass by the number present). If you are referring to a single molecule or ion the mass is in amu’s and is the formula mass. If you are referring to one mole (6.02 × 1023) of the molecule or ions then the mass is in grams and is the molar mass. Mole conversions follow basic dimensional analysis rules and are part of nearly all meaningful stoichiometric calculations.

Example 1 - Particles Determine the number of oxygen atoms in 3.65mol of oxygen gas (O2).

Example 2 - Mass How many moles of iron(II) chloride (FeCl2) do I have if I mass out 45.2g of the substance?

Example 4 – Determine the mass of 55.8L of xenon dioxide gas at STP. Example 5 – Determine the number of iron atoms in 75g of iron(III) oxide.

Basic Stoichiometry Write the BCE for the chemical reaction. Write down the given quantity and convert its quantity to moles if needed. Use the mole ration from the BCE to covert from moles of the given material to moles of the sought material. Convert the number of moles of the sought material to any other quantity desired using a mole conversion. Does your answer make sense?

EXAMPLE 1 Determine the mass of sodium chloride that can be produced when 125.2L of chlorine gas at STP is reacted with excess sodium metal.

EXAMPLE 1 Determine the mass of sodium chloride that can be produced when 125.2L of chlorine gas at STP is reacted with excess sodium metal. Step 1 2 Na + Cl2  2 NaCl Step 2 𝟏𝟐𝟓.𝟐 𝑳 𝑪 𝒍 𝟐 × 𝟏 𝒎𝒐𝒍 𝑪𝒍 𝟐 𝟐𝟐.𝟒 𝑳 𝑪𝒍 𝟐 Step 3 125.2 𝐿 𝐶 𝑙 2 × 1 𝑚𝑜𝑙 𝐶𝑙 2 22.4 𝐿 𝐶𝑙 2 × 𝟐 𝒎𝒐𝒍 𝑵𝒂𝑪𝒍 𝟏 𝒎𝒐𝒍 𝑪𝒍 𝟐 Step 4 125.2 𝐿 𝐶 𝑙 2 × 1 𝑚𝑜𝑙 𝐶𝑙 2 22.4 𝐿 𝐶𝑙 2 × 2 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙 1 𝑚𝑜𝑙 𝐶𝑙 2 × 𝟓𝟖.𝟒𝟒 𝒈 𝑵𝒂𝑪𝒍 𝟏 𝒎𝒐𝒍 𝑵𝒂𝑪𝒍 =𝟔𝟓𝟑.𝟑 𝒈 𝑵𝒂𝑪𝒍 Step 5 Yes the answer makes sense…I should be getting a little more than 1058 as an answer!

Limiting and Excess Reactants Convert the known quantities to moles if required. Divide each number of moles by the corresponding coefficient in the BCE to obtain the “reaction multiplier”. The chemical species with the smallest “reaction multiplier” is the limiting reactant. ALL OF THE LIMITING REACTANT IS CONSUMED AND ITS AMOUNT DETERMINES THE AMOUNTS OF ALL OTHER SPECIES USED OR PRODUCED!!! The chemical species with the greatest “reaction multiplier” is the excess reactant. SOME OF THE EXCESS REACTANT REMAINS AFTER THE REACTION IS COMPLETED!!

EXAMPLE 2 314.2L of nitrogen gas at STP is mixed with 80.0g of hydrogen gas in a reactor. This process, known as the Haber process, can be driven to completion by removal of the product (ammonia). Determine the limiting and excess reactants in this process.

EXAMPLE 2 314.2L of nitrogen gas at STP is mixed with 80.0g of hydrogen gas in a reactor. This process, known as the Haber process, can be driven to completion by removal of the product (ammonia). Determine the limiting and excess reactants in this process. BCE: N2(g) + 3 H2(g)  2 NH3 (g) Step 1 314.2𝐿 𝑁 2 × 1𝑚𝑜𝑙 𝑁 2 22.4𝐿 𝑁 2 =14.03𝑚𝑜𝑙 𝑁 2 80.0𝑔 𝐻 2 × 1𝑚𝑜𝑙 𝐻 2 2.016𝑔 𝐻 2 =39.7𝑚𝑜𝑙 𝐻 2 Step 2 𝟏𝟒.𝟎𝟑𝒎𝒐𝒍 𝑵 𝟐 𝟏𝒎𝒐𝒍 𝑵 𝟐 =𝟏𝟒.𝟎𝟑 𝒕𝒊𝒎𝒆𝒔 𝒕𝒉𝒆 𝑩𝑪𝑬 𝟑𝟗.𝟕𝒎𝒐𝒍 𝑯 𝟐 𝟑 𝒎𝒐𝒍 𝑯 𝟐 =𝟏𝟑.𝟐𝟑 𝒕𝒊𝒎𝒆𝒔 𝒕𝒉𝒆 𝑩𝑪𝑬 Completes reaction more times, so is the EXCESS REACTANT Completes reaction fewer times, so is the LIMITING REACTANT

Theoretical and Percent Yields Theoretical Yields ONLY depend on the LIMITING REACTANT! Use basic stoichiometry from limiting reactant to desired product. Percent Yields are affected by many factors. They are calculated similarly to grades, where the theoretical yield is considered 100%. These values can be less than, equal to, or greater than 100% 𝐏𝐞𝐫𝐜𝐞𝐧𝐭 𝐘𝐢𝐞𝐥𝐝= 𝑨𝒄𝒕𝒖𝒂𝒍 𝒀𝒊𝒆𝒍𝒅 𝑻𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 𝒀𝒊𝒆𝒍𝒅 ×𝟏𝟎𝟎

EXAMPLE 3a 10.5g of lithium metal reacts with 25g of red phosphorous (P4) to create lithium phosphide. Determine the Percent yield of the process if 15g of product is produced.

EXAMPLE 3a (alternative method) 10.5g of lithium metal reacts with 25g of red phosphorous (P4) to create lithium phosphide. Determine the Percent yield of the process if 15g of product is produced. Step 1 (BCE): 12 Li(s) + P4(s)  4 Li3P (s) Step 2: Convert both to amount of product formed. 𝟏𝟎.𝟓𝒈 𝑳𝒊× 𝟏 𝒎𝒐𝒍 𝑳𝒊 𝟔.𝟗𝟒𝒈 𝑳𝒊 × 𝟏 𝒎𝒐𝒍 𝑳𝒊 𝟑 𝑷 𝟑 𝒎𝒐𝒍 𝑳𝒊 × 𝟓𝟏.𝟕𝟗𝒈 𝑳𝒊 𝟑 𝑷 𝟏 𝒎𝒐𝒍 𝑳𝒊 𝟑 𝑷 =𝟐𝟔.𝟏𝟐𝒈 𝑳𝒊 𝟑 𝑷 𝟐𝟓𝒈 𝑷 𝟒 × 𝟏 𝒎𝒐𝒍 𝑷 𝟒 𝟏𝟐𝟑.𝟖𝟖𝒈 𝑷 𝟒 × 𝟒 𝒎𝒐𝒍 𝑳𝒊 𝟑 𝑷 𝟏 𝒎𝒐𝒍 𝑷 𝟒 × 𝟓𝟏.𝟕𝟗𝒈 𝑳𝒊 𝟑 𝑷 𝟏 𝒎𝒐𝒍 𝑳𝒊 𝟑 𝑷 =𝟒𝟐 𝑳𝒊 𝟑 𝑷 Step 3: Determine limiting reactant and theoretical yield. 26.12g Li3P (from 10.5g Li) < 42g Li3 P (from 25g P4) …SO… 26.12g Li3P = theoretical yield Step 4: Find percent yield of product. 𝟏𝟓𝒈 𝑳𝒊 𝟑 𝑷 𝟐𝟔.𝟏𝟐𝒈 𝑳𝒊 𝟑 𝑷 ×𝟏𝟎𝟎=𝟓𝟕%

EXAMPLE 3b 13.6g of hydrogen gas reacts with 200g of chlorine gas to produce hydrogen chloride. Determine the mass of excess reactant remaining if the reaction produces the maximum amount of product possible.

EXAMPLE 3b (another alternative) 13.6g of hydrogen gas reacts with 200g of chlorine gas to produce hydrogen chloride. Determine the mass of excess reactant remaining if the reaction produces the maximum amount of product possible. Step 1 (BCE): H2 (g) + Cl2 (g)  2 HCl (g) Step 2: Choose a reactant and see how much of the other it takes to use it all up. 𝟏𝟑.𝟔𝒈 𝑯 𝟐 × 𝟏 𝒎𝒐𝒍 𝑯 𝟐 𝟐.𝟎𝟐𝒈 𝑯 𝟐 × 𝟏 𝒎𝒐𝒍 𝑪𝒍 𝟐 𝟏 𝒎𝒐𝒍 𝑯 𝟐 × 𝟕𝟏.𝟎𝟎𝒈 𝑪𝒍 𝟐 𝟏 𝒎𝒐𝒍 𝑪𝒍 𝟐 =𝟒𝟕𝟖𝒈 𝑪𝒍 𝟐 𝒏𝒆𝒆𝒅𝒆𝒅 Step 3: Determine limiting reactant. 478g Cl2 needed > 200g available …SO… 200g Cl2 is limiting and H2 is in EXCESS! Step 4: Determine mass of H2 actually used with basic stoichiometry. 𝟐𝟎𝟎𝒈 𝑪𝒍 𝟐 × 𝟏 𝒎𝒐𝒍 𝑪𝒍 𝟐 𝟕𝟏.𝟎𝟎𝒈 𝑪𝒍 𝟐 × 𝟏 𝒎𝒐𝒍 𝑯 𝟐 𝟏 𝒎𝒐𝒍 𝑪𝒍 𝟐 × 𝟐.𝟎𝟐𝒈 𝑯 𝟐 𝟏 𝒎𝒐𝒍 𝑯 𝟐 =5.69g 𝑯 𝟐 𝒖𝒔𝒆𝒅 Step 5: Subtract to see mass of H2 remaining. 𝟏𝟑.𝟔𝒈 𝑯 𝟐 −𝟓.𝟔𝟗𝒈 𝑯 𝟐 =𝟕.𝟗𝟏𝒈 𝑯 𝟐 𝒓𝒆𝒎𝒂𝒊𝒏𝒊𝒏𝒈