Stoichiometry Calculations

Ever wondered how to calculate the exact amount of product in a chemical reaction? Stoichiometry is your answer! Let's break down some examples:

When working with chemical equations like C₁₈H₁₀N₂O₄ + H₂O₂ → Products, you need to use mole ratios from the balanced equation. For melanin oxidation, if 20g of a 9% H₂O₂ solution is used, you'd calculate the mass of H₂O₂ first (1.8g), then convert to moles and apply the mole ratio.

For reactions like 2BF₃ + 3H₂ → 2B + 6HF, identifying the limiting reactant is crucial. Compare the theoretical yields from each reactant - the one producing less product is your limiting reactant. For example, 0.10 mol BF₃ would produce 0.10 mol B, while 0.25 mol H₂ would yield 0.17 mol B, making BF₃ the limiting reactant.

💡 Quick Tip: Always convert to moles first, then apply mole ratios from the balanced equation, and finally convert to the requested unit (mass, volume, etc.).

When calculating product yield for 3C₉H₈ + 5O₂ → 3C₉O₂ + 4H₂O, the same process applies. If you have 3.4g O₂, you'd convert to moles, apply the mole ratio $4 mol H₂O/5 mol O₂$, and then calculate the mass of water produced.

More Limiting Reactant Problems

Chemistry calculations get easier with practice! Let's examine another example with the reaction 3Zn + 2MnO₃ → Mo₂O₃ + 3ZnO.

When given 5.0g Zn and 2.4g MnO₃, you need to determine which reactant will be used up first. Converting to moles:

• Zn: 5.0g × $1 mol/65.40g$ = 0.076 mol
• MnO₃: 2.4g × $1 mol/143.94g$ = 0.025 mol

Next, compare how much product each could theoretically make using the balanced equation's mole ratios. Since MnO₃ would produce less ZnO, it's your limiting reactant, while Zn is in excess.

For reactions involving Mn₃O₄ and Al, you can calculate:

• The number of atoms using Avogadro's number (6.022 × 10²³)
• The number of moles of product using mole ratios

💡 Remember: The limiting reactant is completely consumed in the reaction, while some excess reactant remains unused.

When working with large numbers like 5.33 × 10²⁵ atoms Al, convert to moles first (using Avogadro's number), then apply the appropriate mole ratio from the balanced equation.

Working with Complex Reactions

Ready to tackle more complex chemical calculations? Let's look at calculations for the reaction 2C₅₇H₁₁₀O₆ + 163O₂ → 114CO₂ + 110H₂O.

For gas volume calculations at STP (Standard Temperature and Pressure), remember that 1 mole of any gas occupies 22.4L. If you have 0.64 mol of C₅₇H₁₁₀O₆, you can find the volume of O₂ needed: 0.64 mol × $163 mol O₂/2 mol C₅₇H₁₁₀O₆$ × $22.4L/1 mol O₂$ = 1168L O₂

Similarly, you can calculate the volume of CO₂ produced: 0.64 mol × $114 mol CO₂/2 mol C₅₇H₁₁₀O₆$ × $22.4L/1 mol CO₂$ = 817L CO₂

⚠️ Watch Out: When converting between moles and volume for gases, always check if the conditions are at STP. If not, you'll need to use the ideal gas law $PV=nRT$.

You can also calculate the mass of reactant needed to produce a specific amount of product. For example, to find how many grams of C₅₇H₁₁₀O₆ would produce 55.56 mol H₂O, use the mole ratio and molar mass conversion: 55.56 mol H₂O × $2 mol C₅₇H₁₁₀O₆/110 mol H₂O$ × $891.40g/1 mol C₅₇H₁₁₀O₆$ = 900g C₅₇H₁₁₀O₆.

Chemistry Review and Physics Vectors

Let's look at a complete limiting reactant problem for 3NH₄NO₃ + Na₃PO₄ → (NH₄)₃PO₄ + 3NaNO₃ with 30.0g NH₄NO₃ and 50.0g Na₃PO₄.

To identify the limiting reactant, calculate the theoretical yield of NaNO₃ from each reactant:

• From NH₄NO₃: 50g × $1mol/80.04g$ × $3mol NaNO₃/3mol NH₄NO₃$ × $84.99g/1mol$ = 31.856g NaNO₃
• From Na₃PO₄: 50g × $1mol/164g$ × $3mol NaNO₃/1mol Na₃PO₄$ × $84.99g/1mol$ = 77.74g NaNO₃

Since NH₄NO₃ produces less NaNO₃, it's the limiting reactant. To find the amount of Na₃PO₄ remaining:

• Na₃PO₄ used: 31.856g NH₄NO₃ × $1mol/80.04g$ × $1mol Na₃PO₄/3mol NH₄NO₃$ × $164g/1mol$ = 21.87g
• Na₃PO₄ remaining: 50g - 21.87g = 28.13g

💡 Pro Tip: When calculating remaining reactant, first determine how much was used in the reaction based on the limiting reactant, then subtract from the initial amount.

The physics portion involves vector calculations. When working with direction and magnitude, break each vector into x and y components using sine and cosine, then add all components separately to find the resultant vector.

More Chemistry Practice

Let's practice with 2Mg + O₂ → 2MgO. If we have 2.2g Mg and 4.5L O₂, which is limiting?

For Mg: 2.2g × $1mol/24.3050g$ × $2mol MgO/2mol Mg$ × $22.4L/1mol$ = 2.03L MgO For O₂: 4.5L × $1mol/22.4L$ × $2mol MgO/1mol O₂$ × $22.4L/1mol$ = 9.00L MgO

Since Mg produces less MgO (2.03L vs 9.00L), it's the limiting reactant.

Now for 3CaCO₃ + 2FePO₄ → Ca₃(PO₄)₂ + Fe₂(CO₃)₃ with 100g CaCO₃ and 45g FePO₄:

For CaCO₃: 100g × $1mol/100.09g$ × $1mol Ca₃(PO₄)₂/3mol CaCO₃$ × $310.18g/1mol$ = 103.39g Ca₃(PO₄)₂ For FePO₄: 45g × $1mol/150.82g$ × $1mol Ca₃(PO₄)₂/2mol FePO₄$ × $310.18g/1mol$ = 46.27g Ca₃(PO₄)₂

Since FePO₄ produces less Ca₃(PO₄)₂, it's limiting. To find remaining CaCO₃:

💡 Quick Method: Calculate how much CaCO₃ would react with the limiting reactant, then subtract from the initial amount to find what's left over.

Used CaCO₃: 45g FePO₄ × $1mol/150.82g$ × $3mol CaCO₃/2mol FePO₄$ × $100.09g/1mol$ = 44.06g Remaining CaCO₃: 100g - 44.06g = 55.9g CaCO₃

Percent Yield and Gas Laws

For CuCl₂ + 2NaNO₃ → Cu(NO₃)₂ + 2NaCl with 15g CuCl₂ and 20g NaNO₃:

Calculate theoretical NaCl yield from both reactants:

• From CuCl₂: 15g × $1mol/194.45g$ × $2mol NaCl/1mol CuCl₂$ × $58.44g/1mol$ = 13.04g NaCl
• From NaNO₃: 20g × $1mol/84.99g$ × $2mol NaCl/2mol NaNO₃$ × $58.44g/1mol$ = 13.75g NaCl

Since CuCl₂ produces less NaCl, it's the limiting reactant.

To find unreacted NaNO₃:

• NaNO₃ consumed: 15g CuCl₂ × $1mol/194.45g$ × $2mol NaNO₃/1mol CuCl₂$ × $84.99g/1mol$ = 16.49g
• Remaining NaNO₃: 20g - 16.49g = 3.51g

For percent yield calculation: Percent Yield = $Actual Yield/Theoretical Yield$ × 100 = $11.3g/13.04g$ × 100 = 86.66%

💡 Gas Law Application: For Boyle's Law $P₁V₁ = P₂V₂$, when pressure increases, volume decreases proportionally, and vice versa.

Example: If V₁ = 2.5L at P₁ = 1.5atm, then at P₂ = 1atm: V₂ = (P₁V₁)/P₂ = (1.5atm × 2.5L)/1atm = 3.75L

Molecular and Empirical Formulas

Finding empirical and molecular formulas is a key chemistry skill! Here's the process:

1. Convert mass percentages to moles by dividing by molar mass
2. Find the simplest whole-number ratio by dividing by the smallest value
3. If needed, convert the empirical formula to molecular formula using molar mass

Example 1: With K (39.8g), Mn (27.8g), and O (32.5g):

• Moles: K (1.02 mol), Mn (0.51 mol), O (2.09 mol)
• Divide by smallest (0.51): K (2), Mn (1), O (4)
• Empirical formula: K₂MnO₄

Example 2: With Na (32.4g), S (22.6g), and O (45.0g):

• Moles: Na (1.4 mol), S (0.7 mol), O (2.8 mol)
• Ratio: Na (2), S (1), O (4)
• Empirical formula: Na₂SO₄

💡 Formula Tip: The empirical formula shows the simplest whole-number ratio of atoms, while the molecular formula shows the actual number of atoms in the molecule.

For molecular formulas, determine the empirical formula first, calculate its mass, then find how many empirical formula units make up one molecule:

• Example: C₃H₃O (empirical formula) with molecular weight of 110g
• Empirical formula mass = 55g
• Molecular formula = 2 × empirical formula = C₆H₆O₂

Limiting Reactant and Gas Laws

Let's solve a limiting reactant problem for the combustion reaction 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O:

With 16.4L C₂H₆ and 0.980 mol O₂, determine the limiting reactant by calculating CO₂ produced:

• From C₂H₆: 16.4L × $1mol/22.4L$ × $4mol CO₂/2mol C₂H₆$ × $22.4L/1mol$ = 32.8L CO₂
• From O₂: 0.980 mol × $4mol CO₂/7mol O₂$ × $22.4L/1mol$ = 12.54L CO₂

Since O₂ produces less CO₂, it's the limiting reactant.

To find remaining C₂H₆:

• C₂H₆ consumed: 12.54L CO₂ × $1mol CO₂/22.4L$ × $2mol C₂H₆/4mol CO₂$ = 0.15816 mol
• Initial C₂H₆: 16.4L × $1mol/22.4L$ = 0.73214 mol
• Remaining: 0.73214 - 0.15816 = 0.57398 mol C₂H₆ (or 17.22g)

💡 Gas Law Applications: Know when to use which gas law!

• Boyle's Law $P₁V₁ = P₂V₂$: Relates pressure and volume at constant temperature
• Charles's Law $V₁/T₁ = V₂/T₂$: Relates volume and temperature at constant pressure
• Combined Gas Law: $P₁V₁/T₁ = P₂V₂/T₂$

Example: At 740 torr and 250mL, what volume would gas occupy at 600 torr? P₁V₁ = P₂V₂ (0.97atm)(0.25L) = (0.79atm)(V₂) V₂ = 0.23L or 230mL

Gas Laws Applications

Charles's Law relates volume and temperature at constant pressure. Let's see it in action:

For a gas at 250mL (0.25L) and 25°C (298.15K), what would the volume be at 95°C (368.15K)? V₁/T₁ = V₂/T₂ 0.25L/298.15K = V₂/368.15K V₂ = (0.25L × 368.15K)/298.15K = 0.31L or 308.7mL

The Combined Gas Law relates pressure, volume, and temperature: For a gas at 256mL (0.26L), 720 torr (0.95atm), and 25°C (298.15K), what's the pressure when the volume is 245mL (0.24L) at 50°C (323.15K)? (P₁V₁)/T₁ = (P₂V₂)/T₂ P₂ = (P₁V₁T₂)/(T₁V₂) = (0.95atm × 0.26L × 323.15K)/(298.15K × 0.24L) = 1.12atm or 848 torr

💡 Temperature Conversion: Always convert temperature to Kelvin for gas law calculations by adding 273.15 to the Celsius temperature.

Using Boyle's Law again: If a 3.50L gas at 125kPa (1.23atm) is compressed to 2.00L, the new pressure is: P₂ = (P₁V₁)/V₂ = (1.23atm × 3.50L)/2.00L = 2.15atm or 219kPa

Finally, the Ideal Gas Law $PV = nRT$ can find the number of moles: For 3.9L of gas at 1.00atm and 40°C (313.15K): n = PV/(RT) = (1.00atm × 3.9L)/$0.082058 L·atm/K·mol × 313.15K$ = 0.15117 mol