Chemistry Examples

$C_{14} H_{14} N_{3} O_{3} S N a$

Step 1

Since there is no common factor besides $1$ , the formula is already the empirical formula for this compound.

$C_{14} H_{14} N_{3} O_{3} S N a$

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$