Chemistry - Moles, Molar Mass, and Relative Atomic Mass

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Flashcards covering key definitions, molar mass calculations, relative atomic mass formulas, and mole conversion concepts from the lecture notes.

Last updated 12:33 AM on 9/22/26
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12 Terms

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Relative atomic mass

The weighted average mass of an atom of an element taking into account the mass of all isotopes and their relative abundance.

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Average atomic mass formula (MaveM_{ave})

The formula used to calculate relative atomic mass: Mave=(% abundance×Isotope mass)100M_{ave} = \frac{\sum (\% \text{ abundance} \times \text{Isotope mass})}{100}.

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Avogadro's number

The counting unit constant equal to 6.02×10236.02 \times 10^{23}, representing the number of items in one mole.

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Mole

A unit abbreviated as mol that equals 6.02×10236.02 \times 10^{23} particles, used to measure chemical amounts because individual atoms/molecules are too small to count directly.

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Molar mass

The mass in grams of 1mol1\,\text{mol} of a substance, with units expressed in gmol1\text{g\,mol}^{-1}.

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Atom breakdown of Ca(ClO3)2Ca(ClO_3)_2

A compound formula containing 11 atom of Calcium (CaCa), 22 atoms of Chlorine (ClCl), and 66 atoms of Oxygen (OO).

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Molar mass of Ca(ClO3)2Ca(ClO_3)_2

Calculated as 40.08g40.08\,\text{g} (Ca) + 2(35.45)g2(35.45)\,\text{g} (Cl) + 6(16.00)g6(16.00)\,\text{g} (O) = 206.98g206.98\,\text{g}.

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Atom breakdown of Sn(SO4)2Sn(SO_4)_2

A compound formula containing 11 atom of Tin (SnSn), 22 atoms of Sulfur (SS), and 88 atoms of Oxygen (OO).

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Molar mass of Sn(SO4)2Sn(SO_4)_2

Calculated as 118.71g118.71\,\text{g} (Sn) + 64.12g64.12\,\text{g} (S) + 128g128\,\text{g} (O) = 310.83gmol1310.83\,\text{g\,mol}^{-1}, which rounds to 310.8gmol1310.8\,\text{g\,mol}^{-1}.

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Gold mole conversion

Converting 7.25×108g Au7.25 \times 10^8\,\text{g Au} to moles using molar mass 196.97gmol1196.97\,\text{g\,mol}^{-1} yields 3.68×106mol Au3.68 \times 10^6\,\text{mol Au}.

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Neon relative atomic mass calculation

Calculated as Mave=(20×90.48)+(21×0.27)+(22×9.25)100=1809.6+5.67+203.5100=20.19amuM_{ave} = \frac{(20 \times 90.48) + (21 \times 0.27) + (22 \times 9.25)}{100} = \frac{1809.6 + 5.67 + 203.5}{100} = 20.19\,\text{amu}.

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Lithium relative atomic mass calculation

Calculated as Mave=(6×7.42)+(7×92.58)100=44.52+648.06100=6.9258M_{ave} = \frac{(6 \times 7.42) + (7 \times 92.58)}{100} = \frac{44.52 + 648.06}{100} = 6.9258.