You are watching: What is the difference between molar mass and molecular mass
Also, what"s the difference between molar mass and also molecular mass? Is it simply that molar mass is expressed in Daltons and also molecular mass is expressed in g/mol?
The hardest part around chemisattempt is keeping track of which world are utilizing which terms, and also which terms are outdated. Is this one of these "oh, we use this term now," type of thing?
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edited Sep 29 "15 at 10:22
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Atomic mass refers to the average mass of an atom. This has actually dimensions of mass, so you have the right to express this in regards to daltons, grams, kilograms, pounds (if you really wanted to), or any various other unit of mass. Anyway, as you said, this is an average of the masses of the isotopes, weighted by their loved one abundance. For instance, the atomic mass of $ceO$ is $15.9994~mathrmu$. $mathrmu$ is short for merged atomic mass unit and 1 u is identical to $1.661 imes 10^-24~mathrmg$. It is exactly the exact same as the dalton, yet from what I"ve watched, the term dalton is provided more as soon as stating polymers, biomolecules, or mass spectra.
Molecular mass describes the average mass of a molecule. Again, this has dimensions of mass. It"s simply the sum of the atomic masses of the atoms in a molecule. For example, the molecular mass of $ceO2$ is $2(15.9994~mathrmu) = 31.9988~mathrmu$. You don"t need to calculate the family member isotopic abundance or anything for this bereason it"s currently accounted for in the atomic masses that you are utilizing.
The term molar mass refers to the mass per mole of substance - the name type of means this. This substance can be anything - an facet favor $ceO$, or a molecule prefer $ceO2$. The molar mass has actually units of $mathrmg~mol^-1$, however numerically it is indistinguishable to the two above. So the molar mass of $ceO$ is $15.9994~mathrmg~mol^-1$ and the molar mass of $ceO2$ is $31.9988~mathrmg~mol^-1$.
Sometimes you may come throughout the terms relative atomic mass ($A_mathrmr$) or relative molecular mass ($M_mathrmr$). These are characterized as the ratio of the average mass of one particle (an atom or a molecule) to one-twelfth of the mass of a carbon-12 atom. By interpretation, the carbon-12 atom has a weight of exactly $12~mathrmu$. This is most likely clearer through an instance. Let"s talk about the relative atomic mass of hydrogen, which has actually an atomic mass of $1.008~mathrmu$:$$A_mathrmr(ceH) = frac1.008~mathrmufrac112 imes 12~mathrmu = 1.008$$
Note that this is a proportion of masses and also as such it is dimensionmuch less (it has actually no devices attached to it). But, by meaning, the denominator is constantly equal to $1~mathrmu$ so the relative atomic/molecular mass is constantly numerically equal to the atomic/molecular mass - the just difference is the absence of systems. For example, the family member atomic mass of $ceO$ is 15.9994. The loved one molecular mass of $ceO2$ is 31.9988.
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So in the finish, whatever is numerically the exact same - if you use the correct units - $mathrmu$ and $mathrmg~mol^-1$. There"s nothing stopping you from using units of $mathrmoz~mmol^-1$, it simply won"t be numerically equivalent anyeven more. Which quantity you usage (mass/molar mass/loved one mass) relies on what you are trying to calculate - dimensional evaluation of your equation comes in incredibly handy below.
Summary:Atomic/molecular mass: devices of massMolar mass: systems of mass per amountRelative atomic/molecular mass: no units
A small (and unessential) note around the interpretation of the $ extu$. It"s identified by the $ce^12C$ atom, which is defined to have actually a mass of specifically $12 ext u$. Now the mole is additionally characterized by the $ce^12C$ atom: $12 ext g$ of $ce^12C$ is defined to contain exactly $1 ext mol$ of $ce^12C$. And we understand that one mole of $ce^12C$ consists of $6.022 imes 10^23$ atoms - we contact this number the Avogadro constant. That implies that $12 ext u$ need to be precisely equal to $(12 ext g)/(6.022 imes 10^23)$, and also therefore,
$$1 ext u = frac1 ext g6.022 imes 10^23 = 1.661 imes 10^-24 ext g.$$