Chemistry Problem Solver

The molar mass of a substance is defined as the mass of one mole of that substance. It is expressed in grams per mole (g/mol). To calculate the molar mass of H2O, we need to know the atomic mass of each element in the molecule.

The atomic mass of hydrogen (H) is 1.008 g/mol and the atomic mass of oxygen (O) is 15.999 g/mol. Therefore, the molar mass of H2O can be calculated as follows:

Molar mass of H2O = (2 x atomic mass of hydrogen) + (1 x atomic mass of oxygen)
= (2 x 1.008 g/mol) + (1 x 15.999 g/mol)
= 2.016 g/mol + 15.999 g/mol
= 18.015 g/mol

Therefore, the molar mass of H2O is 18.015 g/mol.

Explanation:

Molar mass is a measurement of the mass of a substance in grams per mole. It is a useful unit when working with chemical reactions and equations because it allows us to easily measure and calculate the amount of substance present.

In this problem, we are given the chemical formula for water, which is H2O. This means that each molecule of water contains 2 atoms of hydrogen (H) and 1 atom of oxygen (O). To calculate the molar mass of H2O, we need to know the atomic mass of each element.

The atomic mass of an element is the mass of one atom of that element relative to the mass of one atom of carbon-12. It is measured in atomic mass units (amu). The periodic table displays the atomic mass of each element, and we can use this information to calculate the molar mass of a compound.

In our case, we have two atoms of hydrogen, each with an atomic mass of 1.008 amu, and one atom of oxygen with an atomic mass of 15.999 amu. So, to find the molar mass of H2O, we simply add the atomic masses of each element, taking into account the number of atoms present in the compound.

This calculation can be represented as:

Molar mass of H2O = (2 x atomic mass of hydrogen) + (1 x atomic mass of oxygen)
= (2 x 1.008 amu) + (1 x 15.999 amu)
= 2.016 amu + 15.999 amu
= 18.015 amu

To convert the value into grams per mole, we simply multiply by the conversion factor of 1 g/mol = 1 amu. This gives us a molar mass of 18.015 g/mol.

In conclusion, the molar mass of H2O is 18.015 g/mol. This value is important in various applications, such as determining the amount of water produced in a chemical reaction or calculating the amount of a substance needed for a specific reaction.

To calculate the molar mass of H2O (water), we need to sum the molar masses of each element in the compound based on its chemical formula.

The chemical formula for water is H2O, which means it contains 2 hydrogen atoms (H) and 1 oxygen atom (O).

Step 1: Find the atomic masses of hydrogen and oxygen

The atomic masses can be found on the periodic table:

• The atomic mass of hydrogen (H) is approximately 1.01 g/mol.

• The atomic mass of oxygen (O) is approximately 16.00 g/mol.

Step 2: Calculate the molar mass of H2O

Since there are 2 hydrogen atoms and 1 oxygen atom in water, the molar mass of H2O can be calculated as follows:

[ \text{Molar mass of H2O} = (2 \times \text{Atomic mass of H}) + (1 \times \text{Atomic mass of O}) ]

[ \text{Molar mass of H2O} = (2 \times 1.01 , \text{g/mol}) + (1 \times 16.00 , \text{g/mol}) ]

[ \text{Molar mass of H2O} = 2.02 , \text{g/mol} + 16.00 , \text{g/mol} ]

[ \text{Molar mass of H2O} = 18.02 , \text{g/mol} ]

Conclusion

The molar mass of H2O (water) is 18.02 g/mol. This value is essential for various calculations in chemistry, such as determining the number of moles in a given mass of water or the mass of water in a given number of moles.

Answer: c) Buchwald–Hartwig Reaction

1. Statement of the Reaction

Copy

Here Ar = an aryl (or heteroaryl) group; R₂NH = primary or secondary amine (R₂ = alkyl or aryl, R₃ = H, alkyl, or aryl).

2. Why it is the Buchwald–Hartwig Reaction

• Coupling Partners - Aryl (pseudo)halide (Cl, Br, I, OTf) - Amines (R₂NH, primary or secondary)

• Catalyst System - Palladium catalyst (e.g. Pd₂(dba)₃, Pd(OAc)₂) - Bulky, electron-rich phosphine ligands (e.g. BINAP, Xantphos, t-Bu₃P) - Base (e.g. NaOtBu, K₃PO₄, Cs₂CO₃)

• Type of Bond Formed - C–N bond (aryl–amine)

• Mechanistic Outline 1. Oxidative Addition Pd(0) + Ar–X → Ar–Pd(II)–X 2. Amine Coordination & Deprotonation Ar–Pd(II)–X + R₂NH → Ar–Pd(II)–NHR₂ + HX (base removes HX) 3. Reductive Elimination Ar–Pd(II)–NHR₂ → Ar–NHR₂ + Pd(0) (catalyst regenerated)

This three-step cycle (oxidative addition → amine coordination/deprotonation → reductive elimination) is the hallmark of the Buchwald–Hartwig amination.

3. Why the Other Options Do Not Match

• a) Ullmann Reaction - Usually copper-mediated, often high temperature, less tolerant of functional groups; generally older C–N couplings.

• b) Gabriel Phthalimide Synthesis - Converts alkyl halides into primary amines via phthalimide (no aryl–amine bond formation).

• d) Chan–Lam Coupling - Copper-mediated coupling of boronic acids (Ar–B(OH)₂) with amines or alcohols under aerobic conditions. Does not use Pd catalyst or aryl halide.

4. Conclusion

The palladium‐catalyzed cross‐coupling of (hetero)aryl halides with amines to form C–N bonds is known as the Buchwald–Hartwig Reaction.