9701 · 2.1
Relative masses of atoms and molecules flashcards
Revision flashcards for Cambridge 9701 Relative masses of atoms and molecules (syllabus 2.1). Flip, recall, then mark a real past-paper question.
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What is relative atomic mass, $A_r$?
The weighted average mass of an atom of an element, compared to 1/12th of the mass of a single carbon-12 atom. It has no units.
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What is relative isotopic mass?
The mass of a single atom of a specific isotope, compared to 1/12th of the mass of a single carbon-12 atom. It is a number very close to the mass number.
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What is relative molecular mass, $M_r$?
The sum of the relative atomic masses of all the atoms in a simple molecule. It is used for covalently bonded substances and has no units.
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What is relative formula mass, $M_r$?
The sum of the relative atomic masses of the atoms in the formula unit of a compound. It is used for ionic compounds (e.g., NaCl) and giant covalent structures (e.g., SiO₂) and has no units.
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Why is the $A_r$ of chlorine 35.5 and not a whole number?
Because it is a weighted average of its naturally occurring isotopes, primarily chlorine-35 (abundance ~75%) and chlorine-37 (abundance ~25%).
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What is the standard used for all relative mass measurements?
An atom of the carbon-12 isotope ($^{12}$C), which is assigned a mass of exactly 12.
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What is the difference in usage between relative molecular mass and relative formula mass?
Strictly, 'molecular mass' applies to simple molecules (like H₂O), while 'formula mass' applies to the formula unit of giant structures (like the ionic lattice NaCl). However, the symbol $M_r$ is commonly used for both.
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Why do relative masses have no units?
They are ratios. The mass of the substance is divided by the mass of the standard (1/12th of a $^{12}$C atom), so any mass units (like kg or g) cancel out.
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Calculate the $M_r$ of ethanol, C₂H₅OH.
$A_r$ values: C=12.0, H=1.0, O=16.0. $M_r = (2 \times 12.0) + (6 \times 1.0) + (1 \times 16.0) = 24.0 + 6.0 + 16.0 = 46.0$.
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Calculate the $M_r$ of ammonium sulfate, (NH₄)₂SO₄.
$A_r$ values: N=14.0, H=1.0, S=32.1, O=16.0. $M_r = 2 \times (14.0 + 4 \times 1.0) + 32.1 + 4 \times 16.0 = 2 \times 18.0 + 32.1 + 64.0 = 36.0 + 32.1 + 64.0 = 132.1$.