Conversion of grams to milliliters using kg/m3 density

Question

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?

Answer 1

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

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

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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.

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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:

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

4. Substitute the unit with the previously calculated expression
5. Multiply the numeric terms

Calculate the volume

Now the unit-adjusted density is used:

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

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Conversion-factors

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:

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

Answer 2

Formulas

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:

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

Calculation

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.