Show Work To Find Solution Explain In The Text How You Appro

Show Work To Findsolutionexplain In The Texthow You Approached An

Show Work To Findsolutionexplain In The Texthow You Approached An

show work to find solution. explain in the text how you approached and worked through the problem. 1) Determine the mass of each of the following: (a) 2.345 mol LiCl (b) 0.0872 mol acetylene, C2H2 (c) 3.3 à— 10−2 mol Na2 CO3 (d) 1.23 à— 103 mol fructose, C6 H12 O6 (e) 0.5758 mol FeSO4(H2O)) Using the periodic table , identify the heaviest member of each of the following groups : (a) alkali metals . (b) chalcogens. (c) noble gases. (d) alkaline earth metals.

Paper For Above instruction

The task involves two main parts: first, calculating the mass of specified quantities of different compounds, and second, identifying the heaviest members of various chemical groups based on atomic weights. This requires a clear understanding of mole concept, molar mass calculations, and periodic table data.

Part 1: Calculating Mass of Substances

To determine the mass of each compound, I employed the fundamental relation:

Mass = number of moles × molar mass

Using the periodic table, I calculated the molar mass of each compound by summing the atomic masses of constituent elements. The atomic masses are approximately: Li = 6.94 g/mol, Cl = 35.45 g/mol, C = 12.01 g/mol, H = 1.008 g/mol, Na = 22.99 g/mol, O = 16.00 g/mol, Fe = 55.85 g/mol, S = 32.07 g/mol.

(a) 2.345 mol LiCl

The molar mass of LiCl is:

6.94 (Li) + 35.45 (Cl) = 42.39 g/mol

Mass = 2.345 mol × 42.39 g/mol ≈ 99.32 grams

(b) 0.0872 mol acetylene, C2H2

Molar mass of C2H2:

(2 × 12.01) + (2 × 1.008) = 24.02 + 2.016 = 26.036 g/mol

Mass = 0.0872 mol × 26.036 g/mol ≈ 2.27 grams

(c) 3.3 × 10^−2 mol Na2CO3

Molar mass of Na2CO3:

(2 × 22.99) + 12.01 + (3 × 16.00) = 45.98 + 12.01 + 48.00 = 105.99 g/mol

Mass = 0.033 mol × 105.99 g/mol ≈ 3.50 grams

(d) 1.23 × 10^3 mol fructose, C6H12O6

Molar mass of C6H12O6:

(6 × 12.01) + (12 × 1.008) + (6 × 16.00) = 72.06 + 12.096 + 96.00 = 180.156 g/mol

Mass = 1230 mol × 180.156 g/mol ≈ 221,532 grams or approximately 221.53 kg

(e) 0.5758 mol FeSO4(H2O)

Molar mass of FeSO4·H2O:

Fe = 55.85 + S = 32.07 + (4 × 16.00) + (2 × 1.008) = 55.85 + 32.07 + 64.00 + 2.016 = 153.936 g/mol

Mass = 0.5758 mol × 153.936 g/mol ≈ 88.75 grams

Part 2: Identifying the Heaviest Member of Each Group

Using the periodic table, I identified the element with the highest atomic weight within each specified group:

(a) Alkali metals

  • Elements: Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Cesium (Cs), Francium (Fr)
  • The heaviest alkali metal is Francium (Fr) with an atomic weight approximately 223 g/mol.

(b) Chalcogens

  • Elements: Oxygen (O), Sulfur (S), Selenium (Se), Tellurium (Te), Polonium (Po)
  • The heaviest chalcogen is Polonium (Po) with an atomic weight around 209 g/mol.

(c) Noble gases

  • Elements: Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), Radon (Rn)
  • The heaviest noble gas is Radon (Rn) with approximately 222 g/mol.

(d) Alkaline earth metals

  • Elements: Beryllium (Be), Magnesium (Mg), Calcium (Ca), Strontium (Sr), Barium (Ba), Radium (Ra)
  • The heaviest is Radium (Ra) with an atomic weight of about 226 g/mol.

Conclusion

This comprehensive analysis combines stoichiometric calculations with periodic table data to provide concrete answers. The pursuit of accurate molar masses and atomic weights allows for precise mass determinations and insightful understanding of element properties within their respective groups.

References

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