Calculating the Mass of Sucrose (C12H22O11) for a Given Mole Quantity

Calculating the Mass of Sucrose (C12H22O11) for a Given Mole Quantity

Introduction to Mole and Calculations

In chemistry, the mole (mol) is a fundamental unit used to measure the amount of substance. When working with any chemical substance, if you know the quantity in moles, you can find the corresponding mass in grams using the molar mass of the substance. This guide will walk you through the process of calculating the mass of sucrose (C12H22O11) for a given quantity of 0.0120 moles.

Understanding Molar Mass and Its Importance

The molar mass (M) of a substance is the mass of one mole of that substance. It is expressed in grams per mole (g/mol) and can be calculated by summing the atomic masses of the elements that make up the compound. For sucrose (C12H22O11), the molar mass can be determined as follows:

Molar Mass Calculation for Sucrose (C12H22O11)

The atomic weights of the elements are used:

Carbon (C): 12.01 g/mol Hydrogen (H): 1.008 g/mol Oxygen (O): 16.00 g/mol

The molar mass of sucrose is calculated by summing the contributions of each element:

Molar mass of C12H22O11  (12.01 g/mol times; 12)   (1.008 g/mol times; 22)   (16.00 g/mol times; 11)                                                              144.12 g/mol   22.176 g/mol   176.00 g/mol                                                              342.296 g/mol

Calculating the Mass from Moles

Given the molar mass of sucrose, we can now calculate the mass of 0.0120 moles of sucrose using the formula:

mass (g)  moles times; molar mass

Substituting the given values:

mass  0.0120 mol times; 342.296 g/mol      4.107552 g

Therefore, the mass of sucrose needed is approximately 4.11 grams.

Conclusion

Understanding the relationship between moles and mass is crucial in chemistry. By knowing the molar mass of a substance and the quantity in moles, you can accurately determine the amount in grams. In the case of sucrose (C12H22O11), 0.0120 moles of sucrose corresponds to approximately 4.11 grams. This knowledge is essential in various applications, including laboratory experiments, pharmaceuticals, and food production.