Question

the mole can be used as a conversion from the atomic to the macroscopic scale. in order to do this molar mass is needed. calculate the molar mass of each chemical formula. type in the numerical response, such as 32.56. round the molar mass of each chemical to two decimal places. ca(c₂h₃o₂)₂

Explanation:

Step1: Identify atomic masses

Atomic masses (from periodic table): Ca = 40.08 g/mol, C = 12.01 g/mol, H = 1.008 g/mol, O = 16.00 g/mol.

Step2: Determine number of each atom

In \( Ca(C_2H_3O_2)_2 \):
- Ca: 1 atom
- C: \( 2 \times 2 = 4 \) atoms (since there are 2 \( C_2H_3O_2 \) groups)
- H: \( 3 \times 2 = 6 \) atoms
- O: \( 2 \times 2 = 4 \) atoms

Step3: Calculate total mass for each element

- Mass of Ca: \( 1 \times 40.08 = 40.08 \) g/mol
- Mass of C: \( 4 \times 12.01 = 48.04 \) g/mol
- Mass of H: \( 6 \times 1.008 = 6.048 \) g/mol
- Mass of O: \( 4 \times 16.00 = 64.00 \) g/mol

Step4: Sum the masses

Molar mass = \( 40.08 + 48.04 + 6.048 + 64.00 \)
= \( 40.08 + 48.04 = 88.12 \); \( 88.12 + 6.048 = 94.168 \); \( 94.168 + 64.00 = 158.168 \)

Step5: Round to two decimal places

\( 158.168 \approx 158.17 \) (Wait, correction: Wait, recalculate the number of atoms again. Wait, \( (C_2H_3O_2)_2 \): C is 2*2=4? Wait no, \( C_2 \) in each group, so two groups: 2*2=4? Wait, no, \( (C_2H_3O_2)_2 \): C: 2 per group, 2 groups: 2*2=4? Wait, no, original formula: \( Ca(C_2H_3O_2)_2 \). Let's re-express the formula: \( CaC_4H_6O_4 \). Wait, no: \( (C_2H_3O_2)_2 \) means 2*(C2H3O2), so C: 2*2=4, H: 3*2=6, O: 2*2=4. Wait, but when I calculated earlier, the sum was wrong. Wait, let's recalculate:

Ca: 40.08

C: 4*12.01 = 48.04

H: 6*1.008 = 6.048

O: 4*16.00 = 64.00

Sum: 40.08 + 48.04 = 88.12; 88.12 + 6.048 = 94.168; 94.168 + 64.00 = 158.168. Wait, but that's incorrect. Wait, no, wait: \( (C_2H_3O_2)_2 \): the subscript 2 applies to all inside the parentheses. So C: 2*2=4, H: 3*2=6, O: 2*2=4. Wait, but let's check the molar mass of calcium acetate. The correct molar mass of \( Ca(C_2H_3O_2)_2 \) (calcium acetate) is calculated as:

Ca: 40.08

C: 2*2=4 atoms: 4*12.01=48.04

H: 3*2=6 atoms: 6*1.008=6.048

O: 2*2=4 atoms: 4*16=64

Total: 40.08 + 48.04 + 6.048 + 64 = 40.08 + 48.04 is 88.12; 88.12 + 6.048 is 94.168; 94.168 + 64 is 158.168. Wait, but the actual molar mass of calcium acetate is 158.17 g/mol? Wait, no, wait, maybe I made a mistake in atomic masses. Wait, H is 1.008, but sometimes rounded to 1.01. Let's recalculate with H=1.01:

H: 6*1.01=6.06

Then total: 40.08 + 48.04 + 6.06 + 64 = 40.08+48.04=88.12; 88.12+6.06=94.18; 94.18+64=158.18. Wait, but standard molar mass of calcium acetate (\( Ca(C_2H_3O_2)_2 \)) is 158.17 g/mol (when using H=1.008: 6*1.008=6.048; 40.08+48.04=88.12; 88.12+6.048=94.168; 94.168+64=158.168, which rounds to 158.17? Wait, no, 158.168 rounded to two decimal places is 158.17? Wait, 158.168: the third decimal is 8, so round up the second: 158.17. Wait, but let's check with correct atomic masses:

Ca: 40.078 (often taken as 40.08)

C: 12.011

H: 1.008

O: 15.999

So recalculating:

Ca: 1*40.078 = 40.078

C: 4*12.011 = 48.044

H: 6*1.008 = 6.048

O: 4*15.999 = 63.996

Sum: 40.078 + 48.044 = 88.122; 88.122 + 6.048 = 94.17; 94.17 + 63.996 = 158.166

Rounded to two decimal places: 158.17? Wait, no, 158.166 is 158.17 when rounded to two decimals? Wait, 158.166: the third decimal is 6, so we round the second decimal (6) up by 1: 158.17. Wait, but actually, 158.166 is closer to 158.17? Wait, no: 158.166, the first decimal is 1, second is 6, third is 6. So when rounding to two decimals, look at the third: 6 ≥5, so add 1 to the second decimal: 6 +1=7. So 158.17. Wait, but I think I made a mistake earlier in the number of atoms. Wait, \( (C_2H_3O_2)_2 \): C is 2*2=4? Wait, no: \( C_2 \) in each group, so two groups: 2*2=4? Wait, no, the formula is \( Ca(C_2H_3O_2)_2 \), which is calcium diacetate, or calcium acetate. The correct formula is \( Ca(C_2H_3O_2)_2 \), which is \( CaC_4H_6O_4 \). So number of atoms: Ca:1, C:4, H:6, O:4. So the calculation is correct. So the molar mass is approximately 158.17 g/mol. Wait, but let's check a reference: Calcium acetate molar mass is 158.17 g/mol (yes, standard value). So the correct calculation gives 158.17 (when rounded to two decimal places). Wait, but in my initial calculation, I had 158.168, which rounds to 158.17. So the correct molar mass is 158.17.

Wait, no, wait: let's do the calculation again with precise atomic masses:

Ca: 40.078 g/mol (1 atom) → 40.078

C: 12.0107 g/mol (4 atoms) → 4*12.0107 = 48.0428

H: 1.00794 g/mol (6 atoms) → 6*1.00794 = 6.04764

O: 15.9994 g/mol (4 atoms) → 4*15.9994 = 63.9976

Sum: 40.078 + 48.0428 = 88.1208; 88.1208 + 6.04764 = 94.16844; 94.16844 + 63.9976 = 158.16604

Rounded to two decimal places: 158.17 (since the third decimal is 6, which is ≥5, so we round the second decimal (6) up to 7).

Answer:

158.17