Convert 100 grams of a solution to milliliters with a density of $\pu{681.9 kg/m3}$. It also has a density of $5.69$ pounds per gallon.

Are any or all of these correct?

\begin{align} \pu{100 g} \times \frac{\pu{1 mL}}{\pu{681.9 g//cm3}} &= \pu{0.146649~mL} &&(\approx \pu{0.15 mL})\\ \pu{1,000 g} \times \frac{\pu{1,000 mL}}{\pu{681.9 g//cm3}} &= \pu{1,466.4906 mL} &&(\approx \pu{1,500 mL})\\ \pu{1 kg} \times \frac{\pu{1 L}}{\pu{681.9 kg//m3}} &= \pu{1.4664906 L} &&(\approx\pu{1.5 L}) \end{align}

Or, can it be as simple as: $$\pu{100 g} \times 681.9 = \pu{68.190 mL}?$$ The unit $\pu{kg/m3}$ is what's confusing me. I'm used to $\pu{g/cm3}$ or $\pu{g/mL}$ for densities. Does $\pu{681.9 g/cm3}$ = $\pu{681.9 kg/m3}$, while scaled-up a thousand-fold? Can I consider using the same density with the math involved by substituting $\pu{kg/m3}$ for $\pu{g/cm3}$ in the equation, or do I have to utilize a conversion within the metric system?


2 Answers 2


Concerning the units in your equation none of them is correct (As all your formulations would end up in a square volume unit).

Definition of volumetric mass density

The density $ \rho $ is given as

$$ \rho = \frac{m}{V} $$

resulting in a mass and density dependent Volume as

$$ V = \frac{m}{\rho} $$

Direct way

Although you could calculate this by just substituting the physical quantities by the given values and units, and then take care of the unit conversion afterward:

\begin{align} V &= \frac{m}{\rho} \\ &= \frac{100\,\mathrm{g}}{681.9\,\mathrm{kg}\,\mathrm{m^{-3}}} =\\ &= \frac{100\,\mathrm{g}\,\mathrm{m^{3}}}{681.9\,\mathrm{kg}} = \frac{100\,\mathrm{g} \cdot \left(10^2\,\mathrm{cm}\right)^{3}}{681.9 \cdot (10^3\,\mathrm{g})} =\\ &= \frac{100 \cdot 10^6\,\mathrm{cm}^{3}}{681.9 \cdot 10^3} = \frac{100 \cdot 10^3}{681.9}\,\mathrm{cm}^{3} =\\ &\approx 146.65\,\mathrm{cm}^{3} \equiv 146.65\,\mathrm{ml} \end{align}

I would suggest to convert the density from $\mathrm{kg}\,\mathrm{m^{-3}}$ to $\mathrm{g}\,\mathrm{cm^{-3}}$ as a first step.

Easier way

Of course easiness is something subjective, but I believe this is easier to use.

Calculate conversion-factor for density

First the combined conversion-factor for the units-fraction is calculated

$$ \frac{\mathrm{kg}}{\mathrm{m}^3} \stackrel{\left[1\right]}{=} \frac{\left(10^3\,\mathrm{g}\right)}{\left(10^2\,\mathrm{cm}\right)^3} \stackrel{\left[2\right]}{=} \frac{10^3\,\mathrm{g}}{10^6\,\mathrm{cm}^3} \stackrel{\left[3\right]}{=} 10^{-3}\,\frac{\mathrm{g}}{\mathrm{cm}^3} $$

  1. Start with substituting each units SI prefix:
  2. Care has to be taken with the conversion of $ \mathrm{m}^3 $, since the power applies to the whole substituted expression ($ \mathrm{m}^3 = \left(10^2\,\mathrm{cm}\right)^3 $).
  3. In the last step the exponential expressions are summarized ($ \frac{10^3}{10^6} = 10^{3-6} = 10^{-3} $).

Then calculate the new numeric value of the density in the target unit:

$$ \rho = 681.9\,\mathrm{kg}\,\mathrm{m^{-3}} \stackrel{\left[4\right]}{\equiv} 681.9 \cdot \left(10^{-3}\,\mathrm{g}\,\mathrm{cm}^{-3}\right) \stackrel{\left[5\right]}{=} 0.6819\,\mathrm{g}\,\mathrm{cm}^{-3} $$

  1. Substitute the unit with the previously calculated expression
  2. Multiply the numeric terms

Calculate the volume

Now the unit-adjusted density is used:

$$ V = \frac{m}{\rho} = \frac{100\,\mathrm{g}}{0.6819\,\mathrm{g}\,\mathrm{cm^{-3}}} = \frac{100}{0.6819}\,\mathrm{cm^{3}} \approx 146.65\,\mathrm{cm}^{3} \equiv 146.65\,\mathrm{ml} $$


According to SI Prefixes the following applies:

\begin{align} 1\,\mathrm{m} &\equiv 10^2\,\mathrm{cm}\\ 1\,\mathrm{kg} &\equiv 10^3\,\mathrm{g} \end{align}

And of course:

$$ 1\,\mathrm{cm}^3 \equiv 1\,\mathrm{ml} $$



Density is defined as mass divided by volume: $ρ = \displaystyle\frac{m}{V}$,
where $ρ$ is density, $m$ is mass, and $V$ is volume.

Your question has:

  • Given $m = 100\text{ }\text{g}$,
  • Given $ρ = 681.9\text{ }\displaystyle\frac{\text{kg}}{\text{m}^3}$,
  • We want to solve for $V$ in units of $\text{mL}$.

By rearranging the density formula, we have: $V = \displaystyle\frac{m}{ρ}$.

Unit conversions

  • $1\text{ }\text{kg} = 1000\text{ }\text{g}$.
  • $1\text{ }\text{m} = 100\text{ }\text{cm}$.
  • $1\text{ }(\text{cm})^3 = 1\text{ }\text{mL}$.

We can rearrange each unit conversion to become a ratio that equals one. That is to say, whenever we have $x = y$, we can say that $\displaystyle\frac{x}{y} = 1 = \displaystyle\frac{y}{x}$. Therefore:

$$ 1 \:=\: \frac{1\text{ }\text{kg}}{1000\text{ }\text{g}} \:=\: \frac{100\text{ }\text{cm}}{1\text{ }\text{m}} \:=\: \frac{1\text{ }\text{mL}}{1\text{ }(\text{cm})^3} $$


We want to compute $V = \displaystyle\frac{m}{ρ}$, which simply involves substituting in the known values. But to get the final answer in the units of $\text{mL}$, we will need to use the unit conversions in a strategic way. Because each conversion is equal to $1$, we can multiply the formula by as many ones as we need until we get the result we want. Namely, $V = \displaystyle\frac{m}{ρ} × 1 × 1 × \cdots × 1$. Here is how we do it:

$V = \displaystyle\frac{100\text{ }\text{g}}{681.9\text{ }\frac{\text{kg}}{\text{m}^3}} = \displaystyle\frac{100\text{ }\text{g}⋅\text{m}^3}{681.9\text{ }\text{kg}}$
$= \displaystyle\frac{100\text{ }\text{g}⋅\text{m}^3}{681.9\text{ }\text{kg}} × 1 × 1^3 × 1$
$= \displaystyle\frac{100\text{ }\text{g}⋅\text{m}^3}{681.9\text{ }\text{kg}} × \displaystyle\frac{1\text{ }\text{kg}}{1000\text{ }\text{g}} × \left(\displaystyle\frac{100\text{ }\text{cm}}{1\text{ }\text{m}}\right)^3 × \displaystyle\frac{1\text{ }\text{mL}}{1\text{ }(\text{cm})^3}$
$= \displaystyle\frac{100\text{ }\text{g}⋅\text{m}^3}{681.9\text{ }\text{kg}} × \displaystyle\frac{1\text{ }\text{kg}}{1000\text{ }\text{g}} × \displaystyle\frac{1000000\text{ }(\text{cm})^3}{1\text{ }\text{m}^3} × \displaystyle\frac{1\text{ }\text{mL}}{1\text{ }(\text{cm})^3}$
$\require{cancel} = \displaystyle\frac{100\text{ }\cancel{\text{g}}⋅\cancel{\text{m}^3}}{681.9\text{ }\cancel{\text{kg}}} × \displaystyle\frac{1\text{ }\cancel{\text{kg}}}{1000\text{ }\cancel{\text{g}}} × \displaystyle\frac{1000000\text{ }\cancel{(\text{cm})^3}}{1\text{ }\cancel{\text{m}^3}} × \displaystyle\frac{1\text{ }\text{mL}}{1\text{ }\cancel{(\text{cm})^3}}$
$= \displaystyle\frac{100}{681.9} × \displaystyle\frac{1}{1000} × \displaystyle\frac{1000000}{1} × \displaystyle\frac{1\text{ }\text{mL}}{1}$
$= \displaystyle\frac{100 × 1 × 1000000 × 1\text{ }\text{mL}}{681.9 × 1000 × 1 × 1}$
$= \displaystyle\frac{100000000}{681900}\text{ }\text{mL}$ $= \displaystyle\frac{1000000}{6819}\text{ }\text{mL}$ (exact)
$≈ 146.6\text{ }\text{mL}$. (rounded to appropriate significant figures)

Notice how we chose conversions so that most of the units in the numerators and denominators cancel out.

  • 1
    $\begingroup$ For future reference: On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax). If you want to know more, please have a look here and here. | It is better to use actual display style, e.g. $$ ... $$ or \begin{align} ... \end{align} than forcing it in inline style $\displaystyle ... $ $\endgroup$ Feb 3, 2022 at 23:07
  • $\begingroup$ @Martin-マーチン I am experienced at LaTex, thank you. My choice of formatting was appropriate in the context of the post. $\endgroup$
    – Nayuki
    Feb 4, 2022 at 2:13
  • $\begingroup$ It's just for your convenience. It looks alright in my desktop browser, but that doesn't extend to mobile. Also two \displaystyle in the same MathJax sequence is unnecessary, only the first one takes effect. So you have a lot of overhead and it takes much longer to load. And please keep in mind: while MathJax uses LaTeX syntax, it absolutely is not LaTeX. $\endgroup$ Feb 4, 2022 at 19:52
  • $\begingroup$ I'm not sure what the problem on mobile is... guessing maybe the math formulas are too long and cannot wrap? As for having two \displaystyle in a sequence, it is indeed redundant in the current text, but I noticed that \displaystyle\frac{a}{\frac{b}{c}} shows up differently than \displaystyle\frac{a}{\displaystyle\frac{b}{c}}, which suggests to me that displaystyle is simply not a global setting. $\endgroup$
    – Nayuki
    Feb 5, 2022 at 5:43
  • 1
    $\begingroup$ Well, $$...$$, or (probably better) \begin{align}...\end{align} would at least allow you to control the amount of vertical space between successive lines. That aside, one of the more impactful differences between MathJax and LaTeX is that on Chemistry.SE there are easier ways to typeset quantities with units; you can use $\pu{3.14 J/(K mol)}$ --> $\pu{3.14 J/(K mol)}$, or since you prefer the fractions, $\pu{3.14 J//K mol}$ -> $\pu{3.14 J//K mol}$. mhchem.github.io/MathJax-mhchem $\endgroup$ Feb 8, 2022 at 18:30

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