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Moles, Mass & Gas Calculations in Stoichiometry
Unlock the power of stoichiometry! Learn to calculate moles, mass, and gas quantities in chemical reactions. Master these fundamental concepts for success in chemistry and beyond.

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Now Playing:Stoichiometry and mass and gas calculations – Example 0a
Intros
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  1. Using mass and gas volume in stoichiometry calculations
  2. Using mass and gas volume in stoichiometry calculations
    Recap stoichiometry basics
  3. Using mass and gas volume in stoichiometry calculations
    Calculating moles
Examples
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  1. Calculate the masses and volumes of reactants and products used in chemical reactions.
    Consider the reaction:

    2C6_6H14  (l)+_{14\;(l)} +19O2  (g)_{2\;(g)} \enspace → \enspace 12CO2  (g)+_{2\;(g)} + 14H2_2O  (g)_{\;(g)}

    1. If 75g of C6_6H14_{14} is burned, what mass of CO2_2 would get produced from this reaction?

    2. If 240g of H2_2O is produced from this reaction, how many moles of C6_6H14_{14} would be required?

    3. At STP, what volume of O2_2 would be required to produce 85 L of CO2_2 in this process?

    4. What mass of C6_6H14_{14} would be required to produce 15 moles of CO2_2?

Introduction to stoichiometry
Jump to:NotesConceptFAQsPrerequisites
Notes
In this lesson, we will learn:
  • To recognize the format of stoichiometry test questions and calculations.
  • To recall the molar volume of gas at standard temperature and pressure and its meaning.
  • Methods to calculate number of moles of chemicals in reactions using mass, moles and volume of gas.
  • To apply Avogadro’s law in finding quantities of gas reactant.

Notes:

  • We saw in Stoichiometry 1: Introduction that because atoms are so small, they cannot be physically counted when doing reactions in a regular laboratory. We use mass, in grams, as a measure of the amount of substance we have because this is related to moles, the actual number of atoms or molecules present by molar mass. Every substance has a molar mass, measured in grams per mole which is the mass of one mole (or 6.023*1023 atoms) of that pure substance.
    • For example, one mole of 12C carbon atoms weighs 12g. If you place pure carbon on a weighing balance and it reads as 12g, you have one mole of carbon atoms.

    We use the number of moles to relate amounts of substance in reactions.
    • For example, see the reaction:

      • CH4 (g) + 2O2 (g)  \, →  \, CO2 (g) + 2H2O (g)

      We know that in this reaction, one mole of CH4 reacts with two moles of O2. CH4 has a molar mass of 16 grams per mol (g mol-1) and O2 has a molar mass of 32 grams per mole.
      With this information we can predict that 64 grams (2 moles with a mass of 32 grams per mole) of O2 will be needed to react with 17 grams of CH4.

    This is the important link between the amount of molecules/substance which we can’t directly measure, and the mass of substance which we can measure in a laboratory.

  • The formula: n(mol)=n(mol) = mass(g)MR(gmol)\large \frac{mass(g)}{M_R (\frac{g}{mol})} can be used to calculate the number of moles of a substance when you know the mass, and you can use the periodic table to find atomic or molecular mass MR of that substance. This applies to elements and molecules, for which you can just add up the elemental mass (e.g. NH3 = 14+1+1+1=17).
    With the moles formula above, you can remember it using the units too, like you would in general algebra: ggmol\large\frac{g}{\frac{g}{mol}} = mol mol and the g will cancel out. You are then left with 11mol\large\frac{1}{\frac{1}{mol}} = mol\, mol which cancels for the mol units.

  • While we can use mass to find number of moles, we need another relation for gases because it is often quite impractical to measure the mass of a gas. Gases are much easier to measure by their volume, or the amount of 3d space they occupy.
    • The molar volume of gas at STP, standard temperature and pressure (0°C or 273K, 100 kPa pressure) is 22.4 litres per mole (22.4 L/mol). In other words, one mole of atoms of a pure ideal gas at 0°C will fill 22.4 litres of space.
    • The molar volume of gas at room temperature (25°C, 298K) and pressure is 24 litres per mole (24 L/mol).

    These values for STP and room temperature and pressure are based on Avogadro’s law which is discussed later.

    Do you notice how a higher temperature makes the same amount of gas take up more space? Both temperature and pressure affect the volume of a gas, which is why we always quote molar volume with temperature and pressure. ALWAYS check the conditions of the reaction in your exam question. Don’t confuse the two!

  • Stoichiometry calculations involve unit conversions from one quantity given in the question to an unknown quantity:
    • To get to moles, use the equation and the molar ratios (the stoichiometric ratio) shown.
    • To get to volume, use the molar volume of gas constants.
    • To get to mass, use the atomic/molecular masses shown in the periodic table.
    You can use conversion factors just like in Introduction: Unit conversions in chemistry. For example, see the reaction:

    O2 (g) + 2Mg (s)   \, →  \, 2MgO (s)

    If the reaction took place at standard temperature and pressure using 72.9 g of Mg, the volume of gas to completely react this amount of magnesium would be:

    72.9 g Mg=g \, Mg =1 mol Mg24.3 g Mg∗1 mol O22 mol Mg∗22.4 L O21 mol O2\large \frac{1 \, mol \, Mg} {24.3 \, g \, Mg} * \frac{1 \, mol \, O_{2}} {2 \, mol \, Mg} * \frac{22.4 \, L \, O_{2}} {1 \, mol \, O_{2}} == 33.6 L O2

    Each fraction in the calculation above is a conversion factor: 24.3 grams of Mg metal is “the same” as 1 mole of Mg metal, we are just changing the units of the expression, not the value of it. The third fraction is the molar volume at standard temperature and pressure we quoted above. Using this, we’ve been able to work out the amount of substance needed for a reaction indirectly!

  • As mentioned above, the molar volumes of gas at RTP and STP are based on Avogadro’s law which states that at constant temperature and pressure, equal volumes of any two gases have the same number of moles.
    This can be expressed as:

    • V1n1=V2n2\large \frac{V_{1}} {n_{1}} = \frac{V_{2}} {n_{2}}

    This is an observation of ideal gas behaviour. When we say “ideal gas” we are making some assumptions about real gases which are not perfectly accurate because real gases do deviate slightly. The assumptions are however very accurate at mild conditions like STP and room temperature. For more on this, see the lesson Ideal gas equation and Kinetic Molecular Theory.

  • Worked example: Using Avogadro’s law At STP, 10 L of O2 gas contains 0.45 mol of O2 molecules. If 3.5 moles more O2 gas is added to the container, what is the new volume?

    • We are being asked for volume (V2) so we will be solving for it using the expression above, rearranged for V2:

    V2=V_{2} = V1n1\large \frac{V_{1}} {n_{1}}  ∗ n2 \, * \, n_{2}

    Using this we can apply V1 = 10L, n1 = 0.45 mol and n2 = 3.95 mol, because 3.5 mol has been added to the 0.45 mol already in the container. This makes the expression:

    V2=V_{2} = 100.45 ∗ \large \frac{10} {0.45} \,* \, 3.95 = 87.8 L O2L \,O_{2}
Concept

Introduction to Stoichiometry: Moles, Mass, and Gas Calculations

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. This section provides a concise overview of moles, mass, and gas calculations in stoichiometry. The introduction video serves as a crucial starting point, offering a visual and interactive explanation of these essential concepts. Understanding moles is vital as they form the basis for many chemical calculations. Mass calculations in stoichiometry allow chemists to determine the quantities of substances involved in reactions. Gas calculations are equally important, especially when dealing with reactions involving gaseous substances. These calculations are indispensable in various fields of chemistry, from laboratory experiments to industrial processes. By mastering these concepts, students and professionals can accurately predict and analyze chemical reactions, making stoichiometry an invaluable tool in the world of chemistry. The principles learned here will be applied throughout your chemistry studies and future career.

FAQs

Here are some frequently asked questions about moles, mass, and gas calculations in stoichiometry:

1. How do you calculate the number of moles from volume?

To calculate the number of moles from volume, use the formula: n = V / Vm, where n is the number of moles, V is the volume of the gas, and Vm is the molar volume (22.4 L/mol at STP). For solutions, use the formula n = CV, where C is the concentration in mol/L and V is the volume in liters.

2. How do you convert mass to moles?

To convert mass to moles, use the formula: n = m / M, where n is the number of moles, m is the mass in grams, and M is the molar mass in g/mol. First, calculate the molar mass of the substance, then divide the given mass by the molar mass.

3. How do you calculate the moles of a gas?

To calculate the moles of a gas, you can use the ideal gas law: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is temperature in Kelvin. Rearrange the equation to solve for n: n = PV / RT.

4. Is a mole of gas always 22.4 L?

A mole of gas occupies 22.4 L only at standard temperature and pressure (STP), which is 0°C (273.15 K) and 1 atm. Under different conditions, the volume can vary. Use the ideal gas law to calculate the volume of a gas at non-STP conditions.

5. How can we find the mass of a gas?

To find the mass of a gas, first calculate the number of moles using the ideal gas law or volume at STP. Then, multiply the number of moles by the molar mass of the gas. The formula is: mass = (number of moles) × (molar mass). For example, if you have 2 moles of O2, the mass would be 2 mol × 32 g/mol = 64 g.

Prerequisites

Understanding the fundamental concepts that lay the groundwork for "Moles, mass and gas calculations" is crucial for students aiming to master this important area of chemistry. While there are no specific prerequisite topics provided in the given JSON format, it's essential to recognize that this subject builds upon several core chemistry principles.

A solid foundation in basic chemistry is vital for comprehending moles, mass, and gas calculations. Students should be familiar with atomic structure, the periodic table, and chemical formulas. These foundational concepts provide the necessary context for understanding how atoms and molecules interact, which is fundamental to mole calculations.

Additionally, a strong grasp of mathematical skills is indispensable. Proficiency in algebra and unit conversions is particularly important, as these skills are frequently applied when working with moles and performing gas calculations. Students who are comfortable with mathematical problem-solving will find it easier to navigate the quantitative aspects of this topic.

An understanding of the states of matter, especially the gaseous state, is also beneficial. Familiarity with the behavior of gases and the factors that affect them, such as temperature and pressure, provides a conceptual framework for gas calculations.

Moreover, knowledge of chemical reactions and stoichiometry serves as a bridge to more advanced mole calculations. These concepts help students understand the relationships between reactants and products in chemical equations, which is essential for solving complex problems involving moles and mass.

While not explicitly listed as prerequisites, topics such as significant figures and scientific notation are also relevant. These concepts ensure accuracy in calculations and proper representation of numerical results, which is crucial in scientific work.

By having a firm grasp of these underlying concepts, students will be better equipped to tackle the complexities of moles, mass, and gas calculations. They will be able to make meaningful connections between different aspects of chemistry and develop a more comprehensive understanding of how these calculations apply to real-world scenarios.

In conclusion, while specific prerequisite topics were not provided, it's clear that a broad foundation in basic chemistry and mathematics is essential for success in studying moles, mass, and gas calculations. Students who take the time to reinforce these fundamental concepts will find themselves better prepared to engage with this challenging but rewarding area of chemistry.