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

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

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

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

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

1. $\ce{3 cd_{3}(po_{4})_{2}}$

Explanation:

Step1: Analyze \( \ce{C_6H_{12}O_6} \)

For the formula \( \ce{C_6H_{12}O_6} \), the subscripts give the ratio of elements. So, the ratio of \( \ce{C:H:O} \) is \( 6:12:6 \), which simplifies to \( 1:2:1 \). The coefficient here is 1 (implied), so the total number of atoms is \( 6 + 12 + 6 = 24 \).

Step2: Analyze \( \ce{3Li_3(PO_4)} \)

First, in \( \ce{Li_3(PO_4)} \), the ratio of \( \ce{Li:P:O} \) is \( 3:1:4 \) (from subscripts). The coefficient is 3, so we multiply each subscript by 3. New subscripts: \( \ce{Li} \): \( 3\times3 = 9 \), \( \ce{P} \): \( 1\times3 = 3 \), \( \ce{O} \): \( 4\times3 = 12 \). Ratio becomes \( 9:3:12 \) or \( 3:1:4 \) (simplified). Total atoms: \( 9 + 3 + 12 = 24 \), and with coefficient 3, total is \( 3\times(3 + 1 + 4)=3\times8 = 24 \)? Wait, no, original in one formula: \( 3 + 1 + 4 = 8 \), times 3 is \( 24 \).

Step3: Analyze \( \ce{2CHCl_3} \)

In \( \ce{CHCl_3} \), ratio \( \ce{C:H:Cl} = 1:1:3 \). Coefficient 2, so multiply each by 2: \( \ce{C} \): 2, \( \ce{H} \): 2, \( \ce{Cl} \): 6. Ratio \( 2:2:6 \) or \( 1:1:3 \). Total atoms per formula: \( 1 + 1 + 3 = 5 \), times 2 is \( 10 \).

Step4: Analyze \( \ce{2Co(CO_3)_3} \)

In \( \ce{Co(CO_3)_3} \), \( \ce{Co} \): 1, \( \ce{C} \): \( 1\times3 = 3 \), \( \ce{O} \): \( 3\times3 = 9 \). Ratio \( \ce{Co:C:O} = 1:3:9 \). Coefficient 2: multiply each by 2. \( \ce{Co} \): 2, \( \ce{C} \): 6, \( \ce{O} \): 18. Ratio \( 2:6:18 \) or \( 1:3:9 \). Total per formula: \( 1 + 3 + 9 = 13 \), times 2 is \( 26 \).

Step5: Analyze \( \ce{4Mn_2(SO_4)_3} \)

In \( \ce{Mn_2(SO_4)_3} \), \( \ce{Mn} \): 2, \( \ce{S} \): \( 1\times3 = 3 \), \( \ce{O} \): \( 4\times3 = 12 \). Ratio \( \ce{Mn:S:O} = 2:3:12 \). Coefficient 4: multiply each by 4. \( \ce{Mn} \): 8, \( \ce{S} \): 12, \( \ce{O} \): 48. Ratio \( 8:12:48 \) or \( 2:3:12 \). Total per formula: \( 2 + 3 + 12 = 17 \), times 4 is \( 68 \).

Step6: Analyze \( \ce{3Cd_3(PO_4)_2} \)

In \( \ce{Cd_3(PO_4)_2} \), \( \ce{Cd} \): 3, \( \ce{P} \): \( 1\times2 = 2 \), \( \ce{O} \): \( 4\times2 = 8 \). Ratio \( \ce{Cd:P:O} = 3:2:8 \). Coefficient 3: multiply each by 3. \( \ce{Cd} \): 9, \( \ce{P} \): 6, \( \ce{O} \): 24. Ratio \( 9:6:24 \) or \( 3:2:8 \). Total per formula: \( 3 + 2 + 8 = 13 \), times 3 is \( 39 \).

Answer:

s (for each formula's element ratio and total atoms):

1. \( \ce{C_6H_{12}O_6} \): Ratio \( \ce{C:H:O} = 1:2:1 \), Total Atoms = 24
2. \( \ce{3Li_3(PO_4)} \): Ratio \( \ce{Li:P:O} = 3:1:4 \), Total Atoms = 24
3. \( \ce{2CHCl_3} \): Ratio \( \ce{C:H:Cl} = 1:1:3 \), Total Atoms = 10
4. \( \ce{2Co(CO_3)_3} \): Ratio \( \ce{Co:C:O} = 1:3:9 \), Total Atoms = 26
5. \( \ce{4Mn_2(SO_4)_3} \): Ratio \( \ce{Mn:S:O} = 2:3:12 \), Total Atoms = 68
6. \( \ce{3Cd_3(PO_4)_2} \): Ratio \( \ce{Cd:P:O} = 3:2:8 \), Total Atoms = 39

(Note: For the "model" part, you can use particle diagrams or table to show subscripts, coefficients, and total atoms. For example, for \( \ce{C_6H_{12}O_6} \), draw 6 C, 12 H, 6 O; for \( \ce{3Li_3(PO_4)} \), draw 3 sets of 3 Li, 1 P, 4 O each, etc.)