2.1.3 Amount of Substance Lecture Notes

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Comprehensive practice flashcards covering mole concepts, calculations, gas laws, and sustainability in chemistry based on the lecture transcript.

Last updated 1:06 PM on 5/25/26
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23 Terms

1
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What is the symbol and unit for amount of substance?

The unit is the mole, and its symbol is mol.

2
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What is the value and definition of the Avogadro constant (NAN_A)?

The number of particles per mole, which is 6.02×1023mol16.02 \times 10^{23}\,mol^{-1}.

3
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Define molar mass and state its typical units.

Molar mass is the mass per mole of a substance, with units of gmol1g\,mol^{-1}.

4
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Define molar gas volume (VmV_m).

The gas volume per mole of gas molecules, with units of dm3mol1dm^3\,mol^{-1} or cm3mol1cm^3\,mol^{-1}.

5
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What is the difference between an empirical formula and a molecular formula?

The empirical formula is the simplest whole number ratio of atoms of each element in a compound, while the molecular formula is the actual number and type of atoms of each element in a molecule.

6
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How many grams are in 1 tonne?

1,000,000g1,000,000\,g (or 1000kg1000\,kg).

7
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What is the conversion factor between dm3dm^3 and cm3cm^3?

1dm3=1000cm31\,dm^3 = 1000\,cm^3.

8
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How do you convert degrees Celsius (C^{\circ}C) into Kelvin (KK)?

0C=273K0^{\circ}C = 273\,K. To convert, add 273273 to the Celsius temperature.

9
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Define the mole based on carbon-12.

The mole is the amount of substance (nn) which contains as many elementary particles as there are atoms in exactly 12g12\,g of carbon-12.

10
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What is the formula to calculate the number of moles (nn) from the number of particles?

n=number of particles6.02×1023n = \frac{\text{number of particles}}{6.02 \times 10^{23}}

11
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What are the elements that exist as diatomic molecules?

H2H_2, N2N_2, O2O_2, F2F_2, Cl2Cl_2, Br2Br_2, I2I_2, and At2At_2.

12
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Define relative atomic mass (ArA_r).

The weighted mean mass of an element compared with 1/12th1/12th of the mass of an atom of carbon-12.

13
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What is the formula for calculating the percentage by mass of an element in a compound?

relative mass of atoms of the element in the compoundtotal relative mass of the compound (Mr)×100\frac{\text{relative mass of atoms of the element in the compound}}{\text{total relative mass of the compound (Mr)}} \times 100

14
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How is the number of moles (nn) calculated from mass (mm) and molar mass (MM)?

n=mMn = \frac{m}{M}

15
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What is the molar gas volume (VmV_m) at room temperature and pressure (RTP)?

Approximately 24.0dm3mol124.0\,dm^3\,mol^{-1} or 24,000cm3mol124,000\,cm^3\,mol^{-1}.

16
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State the Ideal Gas Equation and the units for each variable.

pV=nRTpV = nRT, where pp is pressure in PaPa, VV is volume in m3m^3, nn is moles, RR is the gas constant (8.314Jmol1K18.314\,J\,mol^{-1}\,K^{-1}), and TT is temperature in KK.

17
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How do you calculate the number of moles (nn) for a solution when the volume is given in cm3cm^3?

n=c×V1000n = \frac{c \times V}{1000}, where cc is concentration in moldm3mol\,dm^{-3} and VV is volume in cm3cm^3.

18
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What are the steps to prepare a standard solution from a solid?

  1. Measure mass by difference. 2. Dissolve in a small amount of water. 3. Transfer to a volumetric flask with washings. 4. Fill to the graduation line with distilled water (meniscus). 5. Stopper and invert to mix.
19
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Define percentage uncertainty for a volumetric measurement.

(number of readings×uncertainty of equipmentquantity measured)×100\left( \frac{\text{number of readings} \times \text{uncertainty of equipment}}{\text{quantity measured}} \right) \times 100

20
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What is the formula for percentage yield?

% yield=(actual yieldtheoretical yield)×100\text{\% yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100

21
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Define the term 'limiting reagent'.

The reactant that is not in excess and is used up completely in a reaction, thereby determining the amount of product formed.

22
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What is the formula for percentage atom economy?

% atom economy=(sum of molar masses of desired productsum of molar masses of all products)×100\text{\% atom economy} = \left( \frac{\text{sum of molar masses of desired product}}{\text{sum of molar masses of all products}} \right) \times 100

23
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Why do addition reactions have a high atom economy?

They have an atom economy of 100%100\% because only the desired compound is formed and there are no by-products.