# How do I figure out the significant figures from a question that involves a scientific notation to solve?

## E.g. converting 8.6 x 10^22 molecules C_6H_12O_6 into grams I answered 25 g, but it's wasn't right.

Nov 18, 2016

Here's what I got.

#### Explanation:

When dealing with numbers expressed in scientific notation, you go by how many sig figs are used in the mantissa.

$\textcolor{w h i t e}{a a} \textcolor{b l u e}{m} \times {10}^{\textcolor{red}{n} \textcolor{w h i t e}{a} \stackrel{\textcolor{w h i t e}{a a a a a a}}{\leftarrow}} \textcolor{w h i t e}{a \textcolor{b l a c k}{\text{the")acolor(red)("exponent}} a a}$
$\textcolor{w h i t e}{\frac{a}{a} \textcolor{b l a c k}{\uparrow} a a a a}$
$\textcolor{w h i t e}{\textcolor{b l a c k}{\text{the")acolor(blue)("mantissa}} a}$

In your case, the mantissa is equal to $8.6$ and the exponent to $22$. Here you have

$8.6 \to$ two non-zero-digits $=$ two sig figs

Therefore, you know that the answer must be rounded to two sig figs.

That said, you can calculate the number of grams of glucose that contain that many molecules by going to moles first. Use the Avogadro's constant to find the number of moles present in your sample

8.6 * 10^(22) color(red)(cancel(color(black)("molec. C"_6"H"12"O"_6))) * overbrace(("1 mole C"_6"H"_12"O"_6)/(6.022 * 10^(23)color(red)(cancel(color(black)("molec. C"_6"H"12"O"_6)))))^(color(purple)("Avogadro's constant"))

$= {\text{0.1428 moles C"_6"H"_12"O}}_{6}$

Now you can use the molar mass of glucose to find the number of grams present in the sample

0.1428 color(red)(cancel(color(black)("moles C"_6"H"12"O"_6))) * "180.16 g"/(1color(red)(cancel(color(black)("mole C"_6"H"12"O"_6))))

$= \text{ 25.73 g}$

Now, top round this to two sig figs, use the fact that the digit that the third digit from the left is greater than or equal to $5$.

$7 \ge 5$

This tells you that you must round up the second digit from the left, i.e. the $5$ that follows the $2$ becomes a $6$.

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{mass of glucose " = " 26 g}}}}$