Question

you try
determine the ratio of elements in each formula. use a model to show how the ratio and coefficient together change
the total number of atoms.

1. \t$ce{c_{6}h_{12}o_{6}}$

1. \t$3 ce{li_{3}(po_{4})}$

1. \t$2 ce{chcl_{3}}$

1. \t$2 ce{co(co_{2})_{3}}$

1. \t$4 ce{mn_{2}(so_{4})_{3}}$

1. \t$3 ce{cd_{3}(po_{2})_{2}}$

Explanation:

Problem 1: $\boldsymbol{\ce{C6H12O6}}$

Step 1: Identify Element Ratios

In $\ce{C6H12O6}$, the subscripts for $\ce{C}$, $\ce{H}$, and $\ce{O}$ are 6, 12, and 6, respectively. Simplify the ratio by dividing each by their greatest common divisor (6):
$\ce{C:H:O} = \frac{6}{6}:\frac{12}{6}:\frac{6}{6} = 1:2:1$.

Step 2: Total Atoms (No Coefficient)

Total atoms in one molecule: $6 + 12 + 6 = 24$. If a coefficient (e.g., $n$) were applied, total atoms would be $n \times 24$.

Step 1: Inner Formula Ratios

In $\ce{Li5(PO4)}$, $\ce{Li: P: O} = 5:1:4$ (subscripts: 5 for $\ce{Li}$, 1 for $\ce{P}$, 4 for $\ce{O}$).

Step 2: Apply Coefficient (3)

Total atoms per $\ce{Li5(PO4)}$: $5 + 1 + 4 = 10$. With coefficient 3, total atoms: $3 \times 10 = 30$.
Element ratios scaled by coefficient remain $5:1:4$ (coefficient affects total count, not the ratio within the formula).

Step 1: Inner Formula Ratios

In $\ce{CHCl3}$, $\ce{C:H:Cl} = 1:1:3$ (subscripts: 1, 1, 3).

Step 2: Apply Coefficient (2)

Total atoms per $\ce{CHCl3}$: $1 + 1 + 3 = 5$. With coefficient 2, total atoms: $2 \times 5 = 10$.
Ratio remains $1:1:3$ (coefficient scales total atoms).

Answer:

Element ratio: $\boldsymbol{\ce{C:H:O = 1:2:1}}$; Total atoms per molecule: 24 (scaled by coefficient).

Problem 2: $\boldsymbol{3\ \ce{Li5(PO4)}}$